Question

prove that sec teeta + tan teeta/sec theta-tan theta = [1+sin teeta/costheta]2


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Answer 1

Answer:

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Answer 2

To prove

$\sf\frac{sec\theta +tan\theta}{sec\theta - tan\theta} = (\frac{1+sin\theta}{cos\theta})^2$

Proof

Taking LHS,

Multiply with (secθ + tanθ) on both numerator and denominator

$\sf\frac{(sec\theta + tan\theta)(sec\theta + tan\theta)}{(sec\theta - tan\theta)(sec\theta + tan\theta)}$

Using, (a - b)(a + b) = a² - b²

$\sf\frac{(sec\theta + tan\theta)^2}{sec^2\theta - tan^2\theta}$

We know that, sec²θ - tan²θ = 1

Hence,

$\sf\frac{(sec\theta + tan\theta)^2}{1}$

We know that secθ = 1/cosθ and tanθ = sinθ/cosθ

Hence, we get

$\sf(\frac{1}{cos\theta} + \frac{sin\theta}{cos|theta})^2$

$\sf(\frac{1+sin\theta}{cos\theta})^2$

= RHS

Hence proved