1.

Sin theta-tan theta= a+1÷a-1 then find cos theta

Answer»

Step-by-step explanation:

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}}


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