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Some Applications Of Trigonometry

Question
CBSEENMA10007726

Choose the correct option and Justify your choice.
left parenthesis 1 plus tanθ plus secθ right parenthesis space left parenthesis 1 plus cotθ minus cosecθ right parenthesis = 
(A) 0      (B) 1     (C) 2      (D) -1

Solution
left parenthesis 1 plus tanθ plus secθ right parenthesis space left parenthesis 1 plus cotθ minus cosecθ right parenthesis
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#6 {main}</pre>
equals space open parentheses fraction numerator cosθ plus sinθ plus 1 over denominator cosθ end fraction close parentheses space open parentheses fraction numerator sinθ plus cosθ minus 1 over denominator sinθ end fraction close parentheses
equals space fraction numerator open curly brackets left parenthesis sin space straight theta plus space cos space straight theta right parenthesis plus 1 close curly brackets space open curly brackets left parenthesis sinθ plus cosθ right parenthesis minus 1 close curly brackets over denominator cosθ. space sinθ end fraction
space equals space fraction numerator left parenthesis sinθ plus cosθ right parenthesis squared minus left parenthesis 1 right parenthesis squared over denominator cosθ. space sinθ end fraction
equals fraction numerator sin squared straight theta plus cos squared straight theta plus 2 sinθ. cosθ minus 1 over denominator sinθ. cosθ end fraction
equals space fraction numerator 1 plus 2 sinθ. space cosθ minus 1 over denominator sinθ. space cosθ end fraction space equals space fraction numerator 2 space sinθ. space space cosθ over denominator sinθ. space cosθ end fraction equals 2
Hence, correct option is (C).

Some More Questions From Some Applications of Trigonometry Chapter