Question

Prove the following trigonometric identities:
1+secθ/secθ=sin²θ/1-cosθ

Answer 1

$\text{Consider,}$

$\displaystyle\frac{sin^2\theta}{1-cos\theta}$

$\text{Using,}\bf\,sin^2\theta=1-cos^2\theta$

$=\displaystyle\frac{1-cos^2\theta}{1-cos\theta}$

$=\displaystyle\frac{1^2-cos^2\theta}{1-cos\theta}$

$\text{Using,}\bf\,a^2-b^2=(a-b)(a+b)$

$=\displaystyle\frac{(1-cos\theta)(1+cos\theta)}{1-cos\theta}$

$=\displaystyle\,1+cos\theta$

$=\displaystyle\,1+\frac{1}{sec\theta}$

$=\displaystyle\,\frac{sec\theta+1}{sec\theta}$

$\therefore\bf\,\displaystyle\,\frac{1+sec\theta}{sec\theta}=\frac{sin^2\theta}{1-cos\theta}$

Find more:

Prove that (tanA + secA ÷ cosecA + cotA)(tanA - secA ÷ cosecA - cotA) = 2(tanA×cosecA - cotA×secA)

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