Some Applications of Trigonometry

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Prove the following identity: - Wired Faculty

Question

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

Solution

fraction numerator 1 over denominator secθ minus tanθ end fraction minus 1 over cosθ space equals space 1 over cosθ minus fraction numerator 1 over denominator secθ plus tanθ end fraction

straight L. straight H. straight S. space equals space fraction numerator 1 over denominator secθ minus tanθ end fraction minus 1 over cosθ space space space space space space space space space space space space space equals space open parentheses fraction numerator 1 over denominator secθ minus tanθ end fraction cross times fraction numerator secθ plus tanθ over denominator secθ plus tanθ end fraction close parentheses minus 1 over cosθ space space space space space space space space space space space space space equals open parentheses fraction numerator secθ plus tanθ over denominator sec squared straight theta minus tan squared straight theta end fraction close parentheses minus secθ space space space space space space space space space space space space space space equals space secθ plus tanθ minus secθ space equals space tanθ

straight R. straight H. straight S. equals space 1 over cosθ minus fraction numerator 1 over denominator secθ plus tanθ end fraction space space space space space space space space space space space space space equals space 1 over cosθ minus open parentheses fraction numerator 1 over denominator secθ plus tanθ end fraction cross times fraction numerator secθ minus tanθ over denominator secθ minus tanθ end fraction 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 equals space secθ minus open parentheses fraction numerator secθ minus tanθ over denominator sec squared straight theta minus tan squared straight theta end fraction 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 equals space secθ space minus left parenthesis secθ minus tanθ 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 space space space space space equals space secθ minus secθ plus tanθ 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 tanθ

Hence, L.H.S. = R.H.S.