गणित Chapter 8 त्रिकोणमिति का परिचय
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    NCERT Solution For Class 10 गणित गणित

    त्रिकोणमिति का परिचय Here is the CBSE गणित Chapter 8 for Class 10 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 10 गणित त्रिकोणमिति का परिचय Chapter 8 NCERT Solutions for Class 10 गणित त्रिकोणमिति का परिचय Chapter 8 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 10 गणित.

    Question 1
    CBSEHHIMAH10010205

    ΔABC में, जिसका कोण B समकोण है, AB = 24 cm और BC = 7 cm है। निम्नलिखित का मान ज्ञात कीजिए:
    (i) sin A cos A,
    (ii) sin C, cos C.

    Solution

    हमे प्राप्त हैं, AB = 24 cm
    BC = 7cm, पाइथागोरस प्रमेय के प्रयोग से,
    AC2 = AB2 + BC2
    = (24 cm)2+ (7 cm)2
    = 576 cm2 + 49 cm2
    = 625 cm2
    So, AC = 25 cm

    अब, 
       (i) sin space straight A space equals space BC over AC space equals space 7 over 25 comma space space space cos space straight A space equals space AB over AC space equals 24 over 25
       (ii) sin space straight C space equals space AB over AC space equals space 24 over 25 comma space space space space cos space straight C space equals space BC over AC equals space 7 over 25

    Question 2
    CBSEHHIMAH10010206

    सलंग्न आकृति में tan p - cot R का मान ज्ञात कीजिए।




    Solution

    मान लीजिए:
    PQ = 12K
    तथा 
    PR = 13K
    पाइथागोरस प्रमेय के प्रयोग से:
    PR2 = PQ2 + QR2
    ⇒(13K)2 = (12K)2 + QR2
    ⇒169K2 = 144K2 + QR2
    ⇒QR2 = 169K2 - 144K2
    ⇒ QR2 = 25K2
    ∴ QR = 5K

    अब,  
    tan space straight P space equals space QR over PQ equals fraction numerator 5 straight K over denominator 12 straight K end fraction equals 5 over 12
cot space straight R space equals space QR over PQ equals fraction numerator 5 straight K over denominator 12 straight K end fraction equals 5 over 12
    अत:
    tanP minus cotR space equals space 5 over 12 minus 5 over 12 equals 0.

    Question 3
    CBSEHHIMAH10010207

    यदि sin space straight A space equals space 3 over 4 तो cosA तथा tanA का मान परिकलित कीजिए।

    Solution
    increment माना कि ABC एक समकोण त्रिभुज है जिसमें angle straight B space equals space 90 degree
    हम जानते हैं: 
    space sin space straight A space equals space 3 over 4 space equals space BC over AC
     Let BC = 3K,  AC = 4K
    यहाँ k एक घनात्मक संख्या हैं। 

    पाइथागोरस प्रमेय के प्रयोग से:
                     space space AC squared space equals space AB squared plus BC squared
    rightwards double arrow        space space space space space space left parenthesis 4 straight K right parenthesis squared space equals space AB squared plus left parenthesis 3 straight K right parenthesis squared
    rightwards double arrow              space 16 straight K squared space equals space AB squared plus 9 straight K squared
    rightwards double arrow              space AB squared space equals space 16 straight K squared minus 9 straight K squared
    rightwards double arrow               AB squared space equals space 7 straight K squared
    rightwards double arrow               AB equals square root of 7 straight K
    अब,  cosA space equals space AB over AC equals fraction numerator square root of 7 straight K over denominator 4 straight K end fraction space equals fraction numerator square root of 7 over denominator 4 end fraction
    और    tanA space equals space BC over AB equals fraction numerator 3 straight K over denominator square root of 7 straight K end fraction equals fraction numerator 3 over denominator square root of 7 end fraction.
     
    Question 4
    CBSEHHIMAH10010208

    यदि 15cot A = 8 हो तो sinA और secA का मान ज्ञात कीजिए।

    Solution
    increment ABC माना कि एक समकोण त्रिभुज है जिसमें angle straight B space equals space 90 degree
    हमें दिया गया हैं:
    15 cot A = 8
    rightwards double arrow              space space cotA space equals space 8 over 15 equals AB over BC

    माना 
    AB = 8K, BC = 15K
    पाइथागोरस प्रमेय के प्रयोग से:
    AC2 = AB2 + BC= (8K)2 + (15K)2
    = 64K2 + 225K2
    = 289K2
    अत;
    AC = 17K
    अब,
    AC = 17K
    sinA equals BC over AC equals fraction numerator 15 straight K over denominator 17 straight K end fraction equals 15 over 17
secA space equals space AC over AB equals fraction numerator 17 straight K over denominator 8 straight K end fraction equals 17 over 8.

    Question 5
    CBSEHHIMAH10010209

    यदि sec space straight theta space equals space 13 over 12 हो तो अन्य त्रिकोणमितीय अनुपात परिकलित कीजिए।

    Solution
    माना कि increment ABC एक समकोण त्रिभुज है जिसमें angle straight B space equals space 90 degree  
    हम जानते हैं:
      secθ space equals space 13 over 12 equals AC over AB
    माना कि AB = 12 K, AC = 13K
    यहाँ K एक  घनात्मक संख्या हैं।

    पाइथागोरस प्रमेय के प्रयोग से:
    AC2 = AB2 + BC2
    ⇒(13K)2 = (12K)2 + BC2
    ⇒169K2 = 144K2 + BC2
    ⇒BC2 = 169K2-144K2
    BC2 = 25K2
    rightwards double arrow BC = 5K
    अब,
    cos space straight theta space equals space AB over AC space equals space fraction numerator 12 straight K over denominator 13 straight K end fraction equals space 12 over 13
tan space straight theta space equals space BC over AB space equals fraction numerator 5 straight K over denominator 12 straight K end fraction space equals 5 over 12
cot space straight theta space equals space AB over BC equals space fraction numerator 12 straight K over denominator 5 straight K end fraction equals space 12 over 5
sin space straight theta space equals space BC over AC space equals space fraction numerator 5 straight K over denominator 13 straight K end fraction equals 5 over 13
तथ ा space space cosec space straight theta space equals space AC over BC equals fraction numerator 13 straight K over denominator 5 straight K end fraction equals 13 over 5


 

    Question 6
    CBSEHHIMAH10010210

    यदि angle straight A space space तथ ा space space angle straight B न्यून कोण हैं। जहां cosA = cosB तो दिखाइए कि angle straight A space equals space angle straight B.

    Solution
    cosA space equals space AC over AB
cosB space equals space BC over AB
    अत: AC over AB space equals space BC over AB
    अर्थात्  AC = BC
    अत:   angle straight A space equals space angle straight B

    समान भुजाओं के सम्मुख कोण समान होते हैं।

    Question 7
    CBSEHHIMAH10010211

    यदि cot θ = 7 over 8 comma तो
    fraction numerator left parenthesis 1 plus sinθ right parenthesis space left parenthesis 1 minus sinθ right parenthesis over denominator left parenthesis 1 plus cosθ right parenthesis space left parenthesis 1 minus cosθ right parenthesis end fraction

    Solution
    माना की increment ABC एक समकोण त्रिभुज है, जिसमें angle straight B space equals space 90 degree space तथ ा space angle ACB space equals space straight theta.
    यहाँ, cotθ space equals space BC over AB space equals space 7 over 8
    BC = 7K
    माना BC = 7K
    AB = 8K
    पाइथागोरस प्रमेय के प्रयोग से:
             space space AC squared space equals space AB squared plus BC squared space equals space left parenthesis 8 straight K right parenthesis squared plus left parenthesis 7 straight K right parenthesis squared
                        equals 64 straight K squared plus 49 straight K squared space equals space 113 straight K squared
    rightwards double arrow         AC space equals space square root of 113 straight K

    sinθ space equals space AB over AC space equals space fraction numerator 8 straight K over denominator square root of 113 straight K end fraction equals fraction numerator 8 over denominator square root of 113 end fraction
cosθ space equals space BC over AC space equals space fraction numerator 7 straight K over denominator square root of 113 straight K end fraction equals fraction numerator 7 over denominator square root of 113 end fraction
     अब,
    fraction numerator left parenthesis 1 plus sinθ right parenthesis space left parenthesis 1 minus sinθ right parenthesis over denominator left parenthesis 1 plus cosθ right parenthesis space left parenthesis 1 minus cosθ right parenthesis end fraction equals fraction numerator left parenthesis 1 minus sin squared straight theta right parenthesis over denominator left parenthesis 1 minus cos squared straight theta right parenthesis end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
                                  space space equals space fraction numerator 1 minus open parentheses begin display style fraction numerator 8 over denominator square root of 113 end fraction end style close parentheses squared over denominator 1 minus open parentheses begin display style fraction numerator 7 over denominator square root of 113 end fraction end style close parentheses squared end fraction equals fraction numerator 1 minus begin display style 64 over 113 end style over denominator 1 minus begin display style 49 over 113 end style end fraction
                                     equals space fraction numerator begin display style fraction numerator 113 minus 64 over denominator 113 end fraction end style over denominator begin display style fraction numerator 113 minus 49 over denominator 113 end fraction end style end fraction equals 49 over 64
    Question 8
    CBSEHHIMAH10010212

     यदि 3 cot A = 4, तो जाँच कीजिए कि
    fraction numerator 1 minus tan squared straight A over denominator 1 plus tan squared straight A end fraction equals cos squared straight A minus sin squared straight A है या नहीं।

    Solution
    माना कि increment ABC एक समकोण त्रिभुज है जिसमें angle straight B space equals space 90 degree.
    3 cot A = 4
    rightwards double arrow                 cotA space equals space 4 over 3 space equals space AB over BC
    माना लीजिए, AB = 4K और BC = 3K

    पाइथागोरस प्रमेय के प्रयोग से:
    AC2 = AB2 + BC2
    = (4K)2 + (3K)2
    = 16K2 + 9K2 = 25K2
    अत: AC = 5K
    tan space straight A space equals space BC over AB space equals space fraction numerator 3 straight K over denominator 4 straight K end fraction space equals space 3 over 4
    अब,  cosA space equals space AB over AC equals fraction numerator 4 straight K over denominator 5 straight K end fraction space equals space 4 over 5
    और  sinA space equals space BC over AC equals fraction numerator 3 straight K over denominator 5 straight K end fraction space equals space 3 over 5
         L.H.S. = fraction numerator 1 minus tan squared straight A over denominator 1 plus tan squared straight A end fraction
                  equals space fraction numerator 1 minus open parentheses begin display style 3 over 4 end style close parentheses squared over denominator 1 plus open parentheses begin display style 3 over 4 end style close parentheses squared end fraction space equals space fraction numerator 1 minus begin display style 9 over 16 end style over denominator 1 plus begin display style 9 over 16 end style end fraction
                     equals space fraction numerator begin display style 7 over 16 end style over denominator begin display style 25 over 16 end style end fraction equals 7 over 25
    अब, 
    R.H.S. = cos squared straight A space minus space sin squared straight A
               equals space open parentheses 4 over 5 close parentheses squared minus open parentheses 3 over 5 close parentheses squared
equals space 16 over 25 minus 9 over 25 space equals space fraction numerator 16 minus 9 over denominator 25 end fraction space equals space 7 over 25
    अत: L.H.S. = R.H.S.
    अथवा 
    fraction numerator 1 minus tan squared straight A over denominator 1 plus tan squared straight A end fraction equals cos squared straight A space minus space sin squared straight A

    Question 9
    CBSEHHIMAH10010213

    त्रिभुज ABC में, जिसका कोण B समकोण है, यदि tanA space equals space fraction numerator 1 over denominator square root of 3 end fraction तो निम्नलिखित के मान ज्ञात कीजिए:
    (i) sin A cos C + cos A sin C 
    (ii) cos A cos C - sin A sin C



    Solution

    हमें प्राप्त है: 
    tanA space equals space fraction numerator 1 over denominator square root of 3 end fraction equals BC over AB
    मान लीजिए: AB space equals space square root of 3 straight K comma space space space BC space equals space straight K
    पाइथागोरस प्रमेय के प्रयोग से:
    AC squared space equals space AB squared plus BC squared
space space space space space space space space space equals space left parenthesis square root of 3 straight K right parenthesis squared plus left parenthesis straight K right parenthesis squared
space space space space space space space space space equals space 3 straight K squared plus straight K squared space equals space 4 straight K squared
    AC = 2K

       space sin space straight A space equals space BC over AC equals fraction numerator straight K over denominator 2 straight K end fraction equals 1 half
space sin space straight C space equals space AB over AC equals fraction numerator square root of 3 straight K over denominator 2 straight K end fraction equals fraction numerator square root of 3 over denominator 2 end fraction
अब comma space cos space straight A space equals space AB over AC space equals space fraction numerator square root of 3 straight K over denominator 2 straight K end fraction equals fraction numerator square root of 3 over denominator 2 end fraction
    और, cos space straight C space equals space BC over AC equals fraction numerator straight K over denominator 2 straight K end fraction equals 1 half
    अब, 
    (i) sin A cos C + cos A sin C
    equals space 1 half cross times 1 half plus fraction numerator square root of 3 over denominator 2 end fraction cross times fraction numerator square root of 3 over denominator 2 end fraction
equals space 1 fourth plus 3 over 4 space equals space 4 over 4 space equals space 1
    (ii) cos A. cos C - sin A. sin C
    equals space fraction numerator square root of 3 over denominator 2 end fraction cross times 1 half minus 1 half cross times fraction numerator square root of 3 over denominator 2 end fraction
equals space fraction numerator square root of 3 over denominator 4 end fraction minus fraction numerator square root of 3 over denominator 4 end fraction space equals space 0.

    Question 10
    CBSEHHIMAH10010214

    ΔPQR में, जिसका कोण Q समकोण है, PR + QR =25 cm तथा PQ = 5 cm है। sinP, cosP और tanP के मान ज्ञात कीजिए।

    Solution

    हमें प्राप्त है ,
    PR + QR = 25 cm ..(i)
    माना कि  PR = x cm
    QR = (25 - x) cm.
    पाइथागोरस प्रमेय के प्रयोग से:
    PR2 = PQ2 + RQ2
    ⇒ x2 = (52) + (25 - x)2
    x2 = 25 + 625 + x2 - 50x
    ⇒ 50x = 650
    ⇒ x = 13 cm
    अब, 
    PR = x = 13 cm,  QR = (25 - x) = 12 cm
    और PQ = 5 cm

    अब,

    sin space straight P space equals space RQ over PR space equals space 12 over 13.
cos space straight P space equals space PQ over PR space equals space 5 over 13.
    और tan space straight P space equals space QR over PQ space equals space 12 over 5.

    Question 11
    CBSEHHIMAH10010215

    बताइए कि निम्नलिखित कथन सत्य हैं या असत्य। कारण सहित अपने उत्तर की पुष्टि कीजिए।

    • tan A का मान सदैव 1 से कम होता है।

    • कोण A के किसी मान के लिए secA space equals space 12 over 5

    • cos A, कोण A के cosecant के लिए प्रयुक्त एक संक्षिप्त रूप है। 

    • cot A, cot तथा A का गुणनफल होता है। 

    • cot A, cot तथा A का गुणनफल होता है। 

    Solution

    (i) असत्य (क्योंकि tan space 0 space equals space 0 comma space space space tan space 30 degree space equals space fraction numerator 1 over denominator square root of 3 end fraction comma space tan space 45 degree space equals fraction numerator 1 over denominator square root of 2 end fraction comma
    tan space 60 degree space equals space square root of 3 space तथ ा space space tan space 90 degree space equals space अपर ि भ ा ष ि त)
    (ii) सत्य (क्योंकि विकर्ण, अन्य भुजा से बड़ा होता है।)
    (iii) असत्य, (cosA कोण A से cosine के लिए प्रयुक्त एक संक्षिप्त रूप है।)
    (iv) असत्य, (यह A का cotangent होता है।)
    (v) असत्य (विकर्ण, अन्य भुजा से बड़ा होता है।)

    Question 12
    CBSEHHIMAH10010216

    निम्न के मान निकालिए:
    sin 60° cos 30° + sin 30° cos 60°

    Solution
    sin 60°. cos 30° + sin 30°. cos 60°
    fraction numerator square root of 3 over denominator 2 end fraction cross times fraction numerator square root of 3 over denominator 2 end fraction plus 1 half cross times 1 half
    equals space open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared plus open parentheses 1 half close parentheses squared
    equals space 3 over 4 plus 1 fourth
equals space fraction numerator 3 plus 1 over denominator 4 end fraction equals 4 over 4 equals 1.
    Question 13
    CBSEHHIMAH10010217

    निम्नलिखित के मान लिखिए:
    2 tan245° + cos230° - sin260°

    Solution
    2 tan245° + cos230° - sin260°
                           open square brackets because space tan space 45 degree space equals space 1 comma space space space cos space 30 degree space equals space fraction numerator square root of 3 over denominator 2 end fraction comma space space sin space 60 degree space equals space fraction numerator square root of 3 over denominator 2 end fraction close square brackets
    equals space 2 left parenthesis 1 right parenthesis squared plus open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared minus open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared
equals space 2 plus 3 over 4 minus 3 over 4 space equals space 2
    Question 14
    CBSEHHIMAH10010218

    निम्नलिखित के मान लिखिए:
    fraction numerator cos space 45 degree over denominator sec space 30 degree space plus space cosec space 30 degree end fraction

    Solution
    fraction numerator cos space 45 degree over denominator sec space 30 degree plus cosec space 30 degree end fraction
                    open square brackets because space cos space 45 degree space equals space fraction numerator 1 over denominator square root of 2 end fraction comma space sec space 30 degree space equals space fraction numerator 2 over denominator square root of 3 end fraction comma space cosec space 30 degree space equals 2 close square brackets
    equals space fraction numerator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style over denominator begin display style fraction numerator 2 over denominator square root of 3 end fraction plus 2 over 1 end style end fraction space equals space fraction numerator begin display style fraction numerator 1 over denominator square root of 2 end fraction end style over denominator begin display style fraction numerator 2 plus 2 square root of 3 over denominator square root of 3 end fraction end style end fraction
    equals space fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator square root of 3 over denominator 2 plus 2 square root of 3 end fraction
equals space fraction numerator square root of 3 over denominator square root of 2 left parenthesis 2 plus 2 square root of 3 right parenthesis end fraction
equals space fraction numerator square root of 3 over denominator 2 square root of 2 plus 2 square root of 6 end fraction cross times fraction numerator 2 square root of 2 minus 2 square root of 6 over denominator 2 square root of 2 minus 2 square root of 6 end fraction
equals space fraction numerator square root of 3 left parenthesis 2 square root of 2 minus 2 square root of 6 right parenthesis over denominator left parenthesis 2 square root of 2 right parenthesis squared minus left parenthesis 2 square root of 6 right parenthesis squared end fraction

    equals space fraction numerator 2 square root of 6 minus 2 square root of 18 over denominator 8 minus 24 end fraction equals fraction numerator 2 square root of 6 minus 6 square root of 2 over denominator negative 16 end fraction
equals space fraction numerator 2 left parenthesis square root of 6 minus 3 square root of 2 right parenthesis over denominator negative 16 end fraction equals fraction numerator 3 square root of 2 minus square root of 6 over denominator 8 end fraction

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    Question 19
    CBSEHHIMAH10010223

    यदि tan space left parenthesis straight A plus straight B right parenthesis space equals space square root of 3 और tan left parenthesis straight A minus straight B right parenthesis space equals space fraction numerator 1 over denominator square root of 3 end fraction0 degree less than straight A plus straight B less or equal than 90 degree semicolon space space straight A greater than straight B तो A और B का मान ज्ञात कीजिए।

    Solution
    tan left parenthesis straight A plus straight B right parenthesis space equals space square root of 3
    tan left parenthesis straight A plus straight B right parenthesis space equals space tan space 60 degree
    ⇒ A + B = 60° ...(i)
    और 
    tan left parenthesis straight A minus straight B right parenthesis space equals space fraction numerator 1 over denominator square root of 3 end fraction
    tan left parenthesis straight A minus straight B right parenthesis space equals space tan space 30 degree
    ⇒ A - B = 30° ...(ii)
    समीकरण (i) तथा (ii) को हल करने पर हमें प्राप्त होता हैं:
    A = 45° and B = 15°.
    Question 25
    CBSEHHIMAH10010229

    दिखाइए कि:
    tan 48° tan 23° tan 42° tan 67° = 1.

    Solution

    L.H.S
    = tan 48° . tan 23°. tan 42° . tan 67°
    = tan 48° . tan 42° . tan 23° . tan 67°
    = tan (90° - 42°). tan 42°. tan (90° - 67°). tan 67°
    = cot 42° . tan 42° . cot 67° . tan 67°
    = 1 x 1 = 1 = RHS.
    L.H.S = R.H.S. सिद्ध हुआ।

    Question 26
    CBSEHHIMAH10010230

    दिखाइए कि:
    cos 38° cos 52° - sin 38° sin 52° = 0.

    Solution

    LHS
    = cos 38° . cos 52° - sin 38° . sin 52°
    = cos (90° - 52°). cos 52° - sin (90° - 52°). sin 52°
    = sin 52° . cos 52° - cos 52°. sin 52°
    = 0 = RHS 
    L.H.S. = R.H.S.सिद्ध हुआ।

    Question 27
    CBSEHHIMAH10010231

    यदि tan 2A = cot (A - 18°), जहाँ एक न्यून कोण है, तो A का मान ज्ञान कीजिए।

    Solution

    tan 2A = cot (A - 18°)
                               [∵ cot (90° - θ) = tan θ]
    ⇒ cot (90° - 2A) = cot (A - 18°)
    ⇒ 90° - 2A = A - 18°
    ⇒ -3A = -108°
    ⇒ A = 36°.

    Question 28
    CBSEHHIMAH10010232

    यदि tan A = cot B, तो सिद्ध कीजिए कि A + B = 90°. 

    Solution
    हमें प्राप्त हैं:
    tan A = cot B
    ⇒ cot (90° - A) = cot B
    ⇒ 90° - A = B
    ⇒ 90° = A + B
    अथवा A + B = 90°
    Question 29
    CBSEHHIMAH10010233

    यदि sec 4A = cosec (A - 20°), जहाँ 4A एक न्यून कोण है, तो A का मान बताओ।

    Solution

    हमें प्राप्त हैं: 
    sec 4A = cosec (A - 20°)
    ⇒ cosec (90° - 4A) = cosec (A - 20°)
                                             [∵ cosec (90° - θ) = sec θ]
    ⇒ 90° - 4A = A - 20°
    ⇒ -5 A = -110°
    ⇒ A = 22°

    Question 30
    CBSEHHIMAH10010234

    यदि A, B और C त्रिभुज ABC के अंत:कोण हों,तो दिखाइए कि:  
              space sin space open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses space equals space cos straight A over 2

    Solution

    त्रिभुज ABC, हमें प्राप्त हैं:
                straight A plus straight B plus straight C space equals space 180 degree
    rightwards double arrow space space space space space straight B plus straight C space equals space 180 degree space minus straight A space space space space space... left parenthesis straight i right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space left square bracket उपर ो क ् त space सम ी करण space क ो space left parenthesis 2 right parenthesis space स े space भ ा ग space करन े space पर space right square bracket space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
rightwards double arrow space space space space fraction numerator space straight B plus straight C over denominator 2 end fraction space equals space fraction numerator 180 degree space minus space straight A over denominator 2 end fraction

    rightwards double arrow space space space space space space space space sin space open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses space equals space sin space open parentheses 90 degree minus straight A over 2 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space sin space left parenthesis 90 degree minus straight theta right parenthesis space equals space cos space straight theta close square brackets
rightwards double arrow space space space space space space space space space sin open parentheses fraction numerator straight B plus straight C over denominator 2 end fraction close parentheses space equals space cos straight A over 2
    Question 31
    CBSEHHIMAH10010235

    sin67° + cos75° को 0° और 45° के बीच के कोणों के त्रिकोणमितीय अनुपातों के पदों में व्यक्त कीजिए।

    Solution

    sin 67° + cos 75°
    = sin (90° - 23°) + cos (90° - 15°)
                                  open square brackets because space sin space left parenthesis 90 degree minus straight theta right parenthesis space equals space cosθ close square brackets
    = cos 23° + sin 15°.

    Question 32
    CBSEHHIMAH10010236
    Question 34
    CBSEHHIMAH10010238

    मान निकालिए: 
    fraction numerator sin squared 63 degree plus sin squared 27 degree over denominator cos squared 17 degree plus cos squared 73 degree end fraction

    Solution
    fraction numerator sin squared left parenthesis 90 degree minus 27 degree right parenthesis plus sin squared 27 degree over denominator cos squared left parenthesis 90 degree minus 73 degree right parenthesis plus cos squared 73 degree end fraction
    equals space fraction numerator cos squared 27 degree plus sin squared 27 degree over denominator sin squared 73 degree plus cos squared 73 degree end fraction
    [∵ sin2A + cos2A = 1]
    equals space 1 over 1 equals 1
    Question 35
    CBSEHHIMAH10010239

    मान निकालिए: 
    sin space 25 degree space cos space 65 degree space plus space cos space 25 degree space sin space 65 degree

    Solution

    sin (90° - 65°). cos 65° + cos (90° - 65°). sin 65°
    = cos65°. cos 65° + sin 65°. sin 65°
    = cos2 65° + sin2 65° [∵ sin2 A + cos2 A = 1]
    = 1.

    Question 39
    CBSEHHIMAH10010243

    सही विकल्प चुनिए और अपने विकल्प की पुष्टि कीजिए:
    space fraction numerator 1 plus tan squared straight A over denominator 1 plus cot squared straight A end fraction  बराबर हैं:



    • sec squared straight A
    • -1

    • cot2A

    • tan2A

    Solution

    D.

    tan2A

    fraction numerator 1 plus tan squared straight A over denominator 1 plus cot squared straight A end fraction space equals space fraction numerator sec squared straight A over denominator cosec squared straight A end fraction
                        equals space fraction numerator begin display style fraction numerator 1 over denominator cos squared straight A end fraction end style over denominator begin display style fraction numerator 1 over denominator sin squared straight A end fraction end style end fraction space equals space fraction numerator 1 over denominator cos squared straight A end fraction cross times fraction numerator sin squared straight A over denominator 1 end fraction
                        equals space fraction numerator sin squared straight A over denominator cos squared straight A end fraction space equals space tan squared straight A.

    अत: विकल्प (D) सही हैं।

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    Question 43
    CBSEHHIMAH10010247

    निम्नलिखित सर्वसमिकाएँ सिद्ध कीजिए, जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित हैं, न्यून कोण है:
    fraction numerator 1 plus secA over denominator secA end fraction equals fraction numerator sin squared straight A over denominator 1 minus cosA end fraction


    Solution
    fraction numerator 1 plus secA over denominator secA end fraction equals fraction numerator sin squared straight A over denominator 1 minus cosA end fraction
    L.H.S. = fraction numerator 1 plus secA over denominator secA end fraction

             equals space fraction numerator 1 plus begin display style 1 over cosA end style over denominator begin display style 1 over cosA end style end fraction space equals space fraction numerator begin display style fraction numerator cosA plus 1 over denominator cosA end fraction end style over denominator begin display style 1 over cosA end style end fraction
              equals space 1 plus cosA
    R.H.S. = fraction numerator sin squared straight A over denominator 1 minus cosA end fraction equals fraction numerator 1 minus cos squared straight A over denominator 1 minus cosA end fraction
                                          open square brackets because space sin squared straight A plus cos squared straight A space equals space 1 close square brackets
                                         open square brackets because space straight a squared minus straight b squared space equals space left parenthesis straight a minus straight b right parenthesis thin space left parenthesis straight a plus straight b right parenthesis close square brackets
               equals space fraction numerator left parenthesis 1 plus cosA right parenthesis thin space left parenthesis 1 minus cosA right parenthesis over denominator left parenthesis 1 minus cosA right parenthesis end fraction
equals space 1 plus space cosA                       
    अत:  L.H.S. = R.H.S. 
    Question 46
    CBSEHHIMAH10010250

    निम्नलिखित सर्वसमिकाएँ सिद्ध कीजिए, जहाँ वे कोण, जिनके लिए व्यंजक परिभाषित हैं, न्यून कोण है:
    (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A

    Solution

    (sin A + cosec A)2 + (cos A + sec A)= 7 + tan2A + cot2 A
    L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
                                                   open square brackets because space left parenthesis straight a plus straight b right parenthesis squared space equals space straight a squared plus straight b squared plus 2 ab close square brackets
    = (sin2 A + cosec2 A + 2 sin A. cosec A + cos2 A + sec2 A + 2 cos A. sec A
    = (sin2 A + cos2 A) + 2 sin A. cosec A + 2 cos A. sec A + cosec2 A + sec2 A
    = 1 + 2 + 2 + (cot2 A + 1) + (tan2 A + 1)
    = 5 + 1 + 1 + cot2 A + tan2 A
    = 7 + tan2 A + cot2 A = R.H.S.
    अत: L.H.S. = R.H.S.

    Question 52
    CBSEHHIMAH10010516

    (Cos267o - sin2 23o) का क्या मान है?

    Solution

    Cos2 67° - sin223°

    cos (90° - θ) = sin θ

    माना θ = 23°

    cos (90° - 23°) = sin 23°

    cos 67° = sin 23°

    ∴ cos267° = sin223°

    ∴ cos267° = sin223° = 0

    Question 53
    CBSEHHIMAH10010539

    यदि 4 tanθ = 3 है, तो  का 4 sin θ - cos θ + 14 sin θ + cos θ - 1मान ज्ञात कीजिए

    Solution

    tan θ = 34 tan2 θ = 916हम जानते हैं कि, sec2 θ  = 1 + tan2 θ sec2θ = 1 + 916 = 2516 cos2 θ = 1625 cos θ = 45हम जानते हैं कि, sin2 θ = 1 -cos2  θ sin2 θ = 1-1625 = 925 sin θ = 35अब,4 sin θ - cos θ + 14 sin θ + cos θ -1 = 4 x 35-45 + 14 x  35+45 - 112-4 + 512 + 4-5 = 1311

    Question 54
    CBSEHHIMAH10010540

    यदि tan 2A = cot (A - 18o), जहाँ 2A एक न्यून कोण है, तो A का मान ज्ञात कीजिए

    Solution

    मान लीजिये,

    tan 2A = cot (A - 18o)

    अब, हम जानते हैं कि,

    tan θ = cot (90o - θ)
    ∴ cot (90o -2A)  - cot (A -180)
    ∴ 900 - 2A = A - 180
    ∴ 3A = 1080
    ∴ A  =1080 / 3  = 360

    ∴ A = 360

    Question 55
    CBSEHHIMAH10010543

    सिद्ध कीजिए की sin A - 2sin3 A2 cos3 A - cos A = tan A

    Solution

    sin A - 2 sin3 A2 cos3 A - cos A = tan AL.H.S = sin A (1- 2 sin2 A)cos A (2 cos2 A -1)हम जानते हैं कि, sin2 A + cos2 A  = 1 = sin A cos A(sin2 A + cos2 A - 2 sin2 A)(2 cos2 A - sin 2 A - cos2 A) = tan A cos2 A - sin 2 Acos2 A  - sin 2 A = tan A= R.H.S hence proved

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