This question is from trigonometry topic – trigonometric identities in which we have been asked to prove that (secθ+tanθ)/(secθ-tanθ) = (secθ+tanθ)² = 1+2tan²θ+2secθtanθ
RS Aggarwal, Class 10, chapter 13A, question no 26(ii).
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
We consider (secθ-tanθ)/(secθ+tanθ)
=(secθ-tanθ)/(secθ+tanθ)×(secθ−tanθ)/(secθ−tanθ)
=(secθ-tanθ)²/(sec²θ-tan²θ)
=(secθ-tanθ)²/1
=(secθ−tanθ)²
=sec²θ+tan²θ−2secθtanθ
=(1+tan²θ)+tan²θ−2secθtanθ (sec²θ=1+tan²θ)
=1−2secθtanθ+2tan²θ.