Trigonometry

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Question
SSCCGLENQA12042263

fraction numerator tan squared straight theta over denominator secθ plus 1 end fraction minus secθ is equal to
  • -1

  • 1

  • 0

  • None of these

Solution

A.

-1

We have,    fraction numerator tan squared straight theta over denominator secθ plus 1 end fraction minus secθ
           equals space fraction numerator sec squared straight theta minus 1 over denominator secθ plus 1 end fraction minus secθ
           equals space fraction numerator left parenthesis secθ plus 1 right parenthesis space left parenthesis secθ minus 1 right parenthesis over denominator left parenthesis secθ plus 1 right parenthesis end fraction minus secθ
           equals secθ minus 1 minus secθ equals negative 1
           

Sponsor Area

Question
SSCCGLENQA12042264

sin to the power of 6 straight theta plus cos to the power of 6 straight theta is equal to 
  • 1 minus 3 sin squared θcos squared straight theta
  • 1 minus 3 sinθ space cosθ
  • 1 plus 3 sin squared θcos squared straight theta
  • 1

Solution

A.

1 minus 3 sin squared θcos squared straight theta

We have,  sin to the power of 6 straight theta space plus space cos to the power of 6 straight theta
               equals space left parenthesis sin squared straight theta right parenthesis cubed space plus space left parenthesis cos squared straight theta right parenthesis cubed
               equals left parenthesis sin squared straight theta plus cos squared straight theta right parenthesis space left square bracket sin to the power of 4 straight theta plus cos to the power of 4 straight theta minus sin squared θcos squared straight theta right square bracket
                open square brackets because space straight a cubed plus straight b cubed space equals space left parenthesis straight a plus straight b right parenthesis thin space left parenthesis straight a squared plus straight b squared minus ab right parenthesis close square brackets
               equals space left square bracket left parenthesis sin squared straight theta right parenthesis squared plus left parenthesis cos squared straight theta right parenthesis squared minus sin squared θcos squared straight theta right square bracket
              equals open square brackets open parentheses sin squared straight theta plus cos squared straight theta close parentheses squared minus 2 sin squared straight theta. cos squared straight theta minus sin squared straight theta space cos squared straight theta close square brackets
equals 1 minus 2 sin squared straight theta space cos squared straight theta minus sin squared straight theta space cos squared straight theta
equals 1 minus 3 sin squared θcos squared straight theta
                               
                 

Question
SSCCGLENQA12042810

fraction numerator sinA over denominator 1 plus cosA end fraction space plus space fraction numerator sin begin display style straight A end style over denominator 1 minus cosA end fraction  left parenthesis 0 degree less than space straight A space less than space 90 degree right parenthesis is
  • 2 cosec A

  • 2 sec A

  • 2 sin A

  • 2 cos A

Solution

A.

2 cosec A

fraction numerator sinA over denominator 1 plus cosA end fraction plus fraction numerator sin begin display style straight A end style over denominator 1 minus cosA end fraction
space equals space fraction numerator sinA left parenthesis 1 minus cosA right parenthesis space plus space sin space straight A left parenthesis 1 plus cosA right parenthesis over denominator left parenthesis 1 plus cosA right parenthesis thin space left parenthesis 1 minus cosA right parenthesis end fraction
equals space fraction numerator sinA minus sinA space. space cosA space plus space sinA space plus space sinA space. space cosA over denominator 1 minus cos squared straight A end fraction
equals space fraction numerator 2 space sin space straight A over denominator sin squared straight A end fraction space equals space 2 space cosec space straight A

Question
SSCCGLENQA12042533

(1 + sec 20° + cot 70°)(1 - cosec 20° + tan70°) is equal to

  • 0

  • 1

  • 2

  • 3

Solution

C.

2

left parenthesis 1 plus sec space 20 degree space plus space cot space 70 degree right parenthesis thin space left parenthesis 1 minus cosec space 20 degree space plus space tan space 70 degree right parenthesis
space equals space left parenthesis 1 plus sec space 20 degree space plus space tan space 20 degree right parenthesis thin space left parenthesis 1 minus cosec space 20 degree space plus space cot space 20 degree right parenthesis
open square brackets because space space tan space left parenthesis 90 degree space minus space straight theta right parenthesis space equals space cot space straight theta semicolon space space cot left parenthesis 90 degree space minus space straight theta right parenthesis space equals space tan space straight theta close square brackets
equals space open parentheses 1 plus fraction numerator 1 over denominator cos space 20 degree end fraction space plus space fraction numerator sin space 20 degree over denominator cos space 20 degree end fraction close parentheses space open parentheses 1 minus fraction numerator 1 over denominator sin space 20 degree end fraction plus fraction numerator cos space 20 degree over denominator sin space 20 degree end fraction close parentheses
equals space open parentheses fraction numerator cos space 20 degree plus 1 space plus space sin space 20 degree over denominator cos space 20 degree end fraction close parentheses cross times open parentheses fraction numerator sin space 20 degree space minus space 1 space plus cos space 20 degree over denominator sin space 20 degree end fraction close parentheses
equals space fraction numerator left parenthesis cos space 20 degree space plus space sin space 20 degree right parenthesis squared minus 1 over denominator sin space 20 degree. space cos 20 degree end fraction
equals space fraction numerator cos squared 20 degree space plus space sin squared 20 degree plus 2. space sin 20 degree. space cos 20 degree minus 1 over denominator sin space 20 degree. space cos space 20 degree end fraction
equals space 2

Question
SSCCGLENQA12042613

a, b, c are the lengths of three sides of a triangle ABC. If a, b, c are related by the relation a2 + b2 + c2 = ab + bc + ca, then the value of sin2A+ sin2B + sin2C is

  • 3 over 4
  • fraction numerator 3 square root of 3 over denominator 2 end fraction
  • 3 over 2
  • 9 over 4

Solution

D.

9 over 4 straight a squared plus straight b squared plus straight c squared space equals space ab space plus space bc space plus space ca
rightwards double arrow space space 2 straight a squared space plus space 2 straight b squared space plus space 2 straight c squared space space equals space 2 ab space plus space 2 bc space plus space 2 ca
rightwards double arrow space straight a squared plus space straight a squared space plus space straight b squared space plus space straight b squared space plus space straight c squared space plus space straight c squared space minus space 2 ab space minus space 2 bc minus space 2 ca space equals space 0
rightwards double arrow space space straight a squared plus straight b squared minus 2 ab space plus space straight b squared plus straight c squared minus 2 bc space plus space straight c squared space plus space straight a squared space minus space 2 ca space equals space 0
rightwards double arrow space left parenthesis straight a minus straight b right parenthesis squared space plus space left parenthesis straight b minus straight c right parenthesis squared space plus space left parenthesis straight c minus straight a right parenthesis squared space equals space 0
rightwards double arrow space space space straight a space minus space straight b space equals space 0 space space space rightwards double arrow space space straight a space equals space straight b

     straight b minus straight c space equals space 0 space space space rightwards double arrow space space straight b space equals space straight c
straight c space minus space straight a space equals space 0 space rightwards double arrow space straight c space equals space straight a
Δ ABC is an equilateral triangle.
therefore space space angle straight A space equals space angle straight B space equals space angle straight C space equals space 60 degree
therefore space sin squared straight A space plus space sin squared straight B space plus space sin squared straight C
equals space 3 sin squared straight A space plus space sin squared straight B space plus space sin squared straight C
equals space 3 space sin squared straight A space space equals space 3 space cross times space sin squared 60 degree
equals space 3 space cross times space open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses squared
equals space fraction numerator 3 space cross times space 3 over denominator 4 end fraction space equals space 9 over 4

 

1