Question

If coseco= V2, find the value of
1+sin? 0 + cos² 0
1+sec? o-tan0​

Answer 1

ANSWER:

Given:

• cosec θ = √2

To Find:

• Value of:

$\:\:\:\:\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}$

Solution:

METHOD 1: Using Trigonometric values.

We are given that,

$\implies\cosec\theta=\sqrt2$

So,

$\implies\theta=45^{\circ}$

We need to find:

$\implies\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}$

So,

$\implies\dfrac{1+\sin^2(45^{\circ})+\cos^2(45^{\circ})}{1+\sec^2(45^{\circ})-\tan^2(45^{\circ})}$

$\implies\dfrac{1+\left(\frac{1}{\sqrt2}\right)^2+\left(\frac{1}{\sqrt2}\right)^2}{1+(\sqrt2)^2-(1)^2}$

$\implies\dfrac{1+\frac{1}{2}+\frac{1}{2}}{1+2-1}$

$\implies\dfrac{1+1}{2}$

$\implies\dfrac{2}{2}$

$\implies\bf1$

METHOD 2: Using Trigonometric Identities

We need to find:

$\implies\dfrac{1+\sin^2\theta+\cos^2\theta}{1+\sec^2\theta-\tan^2\theta}$

But, we know that,

$\hookrightarrow \sin^2\phi+\cos^2\phi=1$

And,

$\hookrightarrow \sec^2\phi-\tan^2\phi=1$

So,

$\implies\dfrac{1+(\sin^2\theta+\cos^2\theta)}{1+(\sec^2\theta-\tan^2\theta)}$

$\implies\dfrac{1+1}{1+1}$

$\implies\dfrac{2}{2}$

$\implies\bf1$

Therefore, by both methods the answer is 1.