Question

if tan theta is equal to 1 by root 3 then evaluate cosec square theta minus sec square by cosec square theta plus sec square theta

Answer 1

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Answer 2

Answer:

The expression is $=\frac{\csc^2\theta -\sec^2\theta}{ \csc^2\theta+\sec^2\theta}=\frac{1}{2}$

Step-by-step explanation:

Given : $\tan \theta=\frac{1}{\sqrt3}$

To find : $\frac{\csc^2\theta -\sec^2\theta}{ \csc^2\theta+\sec^2\theta}$

Solution :

Using trigonometry identity,

$\tan \theta=\frac{1}{\sqrt3}=\frac{P}{B}$

Applying Pythagoras theorem,

$H^2=B^2+P^2$

$H^2=\sqrt{3}^2+1^2$

$H^2=3+1$

$H=\sqrt4$

$H=2$

So, P=1 , H=2 ,$B=\sqrt{3}$

We know,

$\csc \theta=\frac{H}{P}$

$\csc \theta=\frac{2}{1}=2$

$\csc^2 \theta=2^2=4$

$\sec \theta=\frac{H}{B}$

$\sec \theta=\frac{2}{\sqrt{3}}$

$\sec^2\theta=(\frac{2}{\sqrt{3}})^2$

$\sec^2\theta=\frac{4}{3}$

Substitute all the values in the expression,

$=\frac{\csc^2\theta -\sec^2\theta}{ \csc^2\theta+\sec^2\theta}$

$=\frac{4 -\frac{4}{3}}{4+\frac{4}{3}}$

$=\frac{\frac{12-4}{3}}{\frac{12+4}{3}}$

$=\frac{\frac{8}{3}}{\frac{16}{3}}$

$=\frac{8\times 3}{3\times 16}$

$=\frac{1}{2}$

Therefore, The expression is $=\frac{\csc^2\theta -\sec^2\theta}{ \csc^2\theta+\sec^2\theta}=\frac{1}{2}$