Question

If secθ + tanθ = p, show that secθ –tanθ = 1/p . Hence, find the values of cosθ and sinθ.

Answer 1

The values of

$\bold{\cos \theta=\frac{2 p}{1+p^{2}}}$

$\bold{\sin\theta=\frac{1-p^{2}}{1+p^{2}}}$

Step-by-step explanation:

From question, the condition given is:

$\bold{\sec \theta+\tan \theta=p\longrightarrow(1)}$

We know that, $\bold{\sec ^{2} \theta-\tan ^{2} \theta=1}$

Since, it is similar to $\bold{a^{2}-b^{2}}$ algebraic expression.

On expanding the above algebraic expression, we get,

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

a = secθ and b = tanθ

On substituting the value of a and b, we get,

$\bold{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=1}$

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

On substituting the given condition in above equation, we get,

$\bold{\sec \theta-\tan \theta=\frac{1}{p}\longrightarrow(2)}$

On adding equation (1) and equation (2), we get,

$\bold{2 \sec \theta=p+\frac{1}{p}}$

$\bold{\frac{1}{\sec \theta}=\frac{2}{p+\frac{1}{p}}}$

$\bold{\therefore\cos \theta=\frac{2 p}{1+p^{2}}}$

We know that, $\bold{\sin \theta=\sqrt{1-\cos ^{2} \theta}}$

$\bold{\sin\theta=\sqrt{1-\frac{4 p^{2}}{1+2 p^{2}+p^{4}}}}$

$\bold{\sin\theta=\sqrt{\frac{1+2 p^{2}-4 p^{2}+p^{4}}{1+2 p^{2}+p^{4}}}}$

$\bold{\sin\theta=\sqrt{\frac{\left(1-p^{2}\right)^{2}}{\left(1+p^{2}\right)^{2}}}}$

$\bold{\therefore\sin\theta=\frac{1-p^{2}}{1+p^{2}}}$