Question

if tan theta = 1/root 7 . find the value of cosec^2 theta - sec^2 theta/cosec^2 theta+sec^2 theta

Answer 1

Tan theta = opposite side/adjacent side
tan theta=1/root 7
Therefore,
Opposite side=1 and Adjacent side= root 7
 By Pythogoras theorem
Hypotenuse^2=Adjacent^2+Opposite^2
Hypotenuse^2 = root 7^2 + 1^2
Hypotenuse^2 = 7+1
Hypotenuse = root 8

Sin theta = Opposite/ Hypotenuse
Sin theta = 1/root 8
Therefore,
cosec theta = root 8/1

cos theta = Adjacent/Hypotenuse
Cos theta = root 7/root 8
Therfore,
sec theta = root 8/ root 7

Now,
Cosec^2 theta - Sec^2 theta / Cosec^2 theta + Sec^2 theta

Root 8^2/1^2 - Root8^2/root 7^2 / root 8^2/1^2 + root8^2/root 7^2

8/1-8/7 / 8/1+8/7

56-8/7 /56+8/7

48/7 / 64/7

48/7 *7/64
 
48/64

3/4

The answer is 3/4. 

Answer 2

"The value will be $\frac { 3 }{ 4 }$

Given:

$\tan \theta=\frac{1}{\sqrt({7} )}$

To find:

The value of $\frac{\csc^{2} \theta-\sec ^{2} \theta}{\csc ^{2} \theta+\sec ^{2} \theta}$

Solution:

$\tan \theta = \frac {\text (opposite side)}{\text (adjacent side)}$

$\tan \theta =\frac{1}{\sqrt({7} )}$

Therefore,

Opposite side = 1 and Adjacent side $= \sqrt({7})$

By Pythogoras theorem

$Hypotenuse ^2 = Adjacent^2+Opposite^2$

$Hypotenuse^2=({\sqrt7})^2 + 1^2$

$Hypotenuse^2 = 7 + 1$

$Hypotenuse = \sqrt 8$

$\Sin \theta = \frac {\text(Opposite)}{\text(Hypotenuse)}$

Therefore,

$\sin \theta=\frac{1}{\sqrt{8}}$

$\csc \theta=\frac{\sqrt{8}}{1}$

$\cos \theta = \frac {\text (Adjacent)}{\text(Hypotenuse)}$

$\cos \theta=\frac{\sqrt{7}}{\sqrt{8}}$

Therefore,

$\sec \theta=\frac{\sqrt{8}}{\sqrt{7}}$

Now,

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

$\frac {\frac{\sqrt{8}^2}{1^2}-\frac{\sqrt{8}^{2}}{\sqrt{7}^{2}}} {\frac{\sqrt{8}^{2}}{1^2}+\frac{\sqrt{8}^{2}}{\sqrt{7}^{2}}}$

$\frac{8-\frac{8}{7}}{8+\frac{8}{7}}$

$\frac{\frac{56-8}{7}}{\frac{56+8}{7}}$

$\frac{48}{7} \times \frac{7}{64}$

$\frac {48}{64}$

$\frac { 3 }{ 4 }$"