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If tan θ=ab, find the value of cos θ+sin θ cos θ−sin θ.

Question:

If $\tan \theta=\frac{a}{b}$, find the value of $\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}$.

Solution:

Given:

$\tan \theta=\frac{a}{b} \ldots \ldots$(1)

Now, we know that $\tan \theta=\frac{\sin \theta}{\cos \theta}$

Therefore equation (1) becomes as follows

$\frac{\sin \theta}{\cos \theta}=\frac{a}{b}$

Now, by applying invertendo

We get,

$\frac{\cos \theta}{\sin \theta}=\frac{b}{a}$

Now, by applying Compenendo-dividendo

We get,

$\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{b+a}{b-a}$

Therefore,

$\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{b+a}{b-a}$

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