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Some Applications Of Trigonometry

Question
CBSEENMA10008279
Wired Faculty App

Prove the following identity:
fraction numerator tanθ plus secθ minus 1 over denominator tanθ minus secθ plus 1 end fraction equals fraction numerator 1 plus sinθ over denominator cosθ end fraction

Solution
straight L. straight H. straight S. space equals space fraction numerator tan space straight theta space plus sec space straight theta space minus 1 over denominator tanθ minus secθ plus 1 end fraction space space space space space space space space space space space space space space equals space fraction numerator left parenthesis tanθ plus secθ right parenthesis space minus space left parenthesis sec squared straight theta minus tan squared straight theta right parenthesis over denominator tanθ minus secθ plus 1 end fraction space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis tanθ plus secθ right parenthesis minus left parenthesis secθ plus tanθ right parenthesis space left parenthesis secθ minus tanθ right parenthesis over denominator tanθ minus secθ plus 1 end fraction space space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis tanθ plus secθ right parenthesis space left square bracket 1 minus left parenthesis secθ plus tanθ right parenthesis over denominator tanθ minus secθ plus 1 end fraction space space space space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis tanθ plus secθ right parenthesis thin space left parenthesis 1 minus secθ plus tanθ right parenthesis over denominator tanθ minus secθ plus 1 end fraction space space space space space space space space space space space space space space space space space space equals space tanθ plus secθ space space space space space space space space space space space space space space space space space space equals space sinθ over cosθ plus 1 over cosθ space space space space space space space space space space space space space space space space space space equals space fraction numerator sinθ plus 1 over denominator cosθ end fraction space equals space straight R. straight H. straight S. space
Hence, L.H.S. = R.H.S.

Some More Questions From Some Applications of Trigonometry Chapter

Choose the correct option and justify your choice:
sin 2A = 2 sin A is true when A = 
(A) 0°         (B) 30°         (C) 45°       (D) 60°

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