Question

If cosec ∅ = √2, find the value of 1+sin²∅+cos²∅/1+sec²∅-tan²∅.​

Answer 1

Given: $cosec$ $A$  $=\sqrt{2}$

∴ $A=\frac{\pi }{4} =45^{0}$

$sin$ $A$ $=sin$ $45^{0}$ = $\frac{1}\sqrt{2}$

$cot$ $A$ $=cot$ $45^{0}$ $=1,$

$tan$ $A$ $=$ $tan$ $45^{0}$ $=1$

$cos$ $A$ $=$ $cot$ $45^{0}$ $=\frac{1}{\sqrt{2} }$

∴ $=2$

Answer 2

$\large\underline{\sf{Solution-}}$

METHOD 1: Using Trigonometric values.

We have given that,

$\sf\cosec\theta=\sqrt2$

So,

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

We need to find:

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

So,

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

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

$\sf\longmapsto\dfrac{1+\frac{1}{2}+\frac{1}{2}}{1+2-1}$

$\sf\longmapsto\dfrac{1+1}{2}$

$\sf\longmapsto\dfrac{2}{2}$

$\sf\longmapsto\bf1$

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,

$\sf\red⇛ \sin^2\phi+\cos^2\phi=1$

And,

$\sf\red⇛ \sec^2\phi-\tan^2\phi=1$

So,

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

$\sf\longmapsto\dfrac{1+1}{1+1}$

$\sf\longmapsto\dfrac{2}{2}$

$\sf\longmapsto\bf1$

Therefore, by both methods the answer is 1.

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