Question

sec theta -tan theta=a+1/a-1 then find cos theta=

Answer 1

Answer:

$\bf\:\cos\theta=\frac{a^2-1}{a^2+1}$

Step-by-step explanation:

Given:

$\sec\theta-\tan\theta=\frac{a+1}{a-1}$........(1)

We know that

$\boxed{\bf\sec^2\theta-\tan^2\theta=1}$

Using

$\boxed{a^2-b^2=(a-b)(a+b)}$

$\implies(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$

$\implies\frac{a+1}{a-1}(\sec\theta+\tan\theta)=1$

$\implies\sec\theta+\tan\theta=\frac{a-1}{a+1}$........(2)

Adding (1) and (2), we get

$2\sec\theta=\frac{a+1}{a-1}+\frac{a-1}{a+1}$

$2\sec\theta=\frac{(a+1)^2+(a-1)^2}{(a-1)(a+1)}$

$2\sec\theta=\frac{a^2+1+2a+a^2+1-2a}{a^2-1}$

$2\sec\theta=\frac{2a^2+2}{a^2-1}$

$2\sec\theta=\frac{2(a^2+1)}{a^2-1}$

Cancelling 2 on both sides,

$\sec\theta=\frac{a^2+1}{a^2-1}$

Taking reciprocals on both sides, we get

$\boxed{\bf\:\cos\theta=\frac{a^2-1}{a^2+1}}$

Answer 2

Answer:

(a^2-1)/(a^2+1)

Step-by-step explanation:

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