Math, asked by shivamji, 1 year ago

sin squared theta + 1 by 1 + tan squared theta

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Answers

Answered by qwmagpies

The value is 1.

Given: Given expression is

${sin}^{2} \alpha + \frac{1}{1 + { \tan}^{2} \alpha }$

To find: We have to simplify it.

Solution:

We know that,

${ \sec }^{2} \alpha - { \tan}^{2} \alpha = 1$

Thus putting the value of tan alpha in terms of sec alpha we get-

${ \sec }^{2} \alpha = { \tan}^{2} \alpha + 1$

Again we know that

${ \sin }^{2} \alpha + { \cos }^{2} \alpha = 1$

Putting the value of tan alpha in the above expression we get-

${sin}^{2} \alpha + \frac{1}{1 + { \tan}^{2} \alpha } \\ {sin}^{2} \alpha + \frac{1}{ { \sec }^{2} \alpha } \\ {sin}^{2} \alpha + { \cos}^{2} \alpha \\ = 1$

Thus the simplified value of the expression is 1.

Answered by pulakmath007

$\displaystyle \sf{ {sin}^{2} \theta + \frac{1}{(1 + {tan}^{2} \theta) } = 1}$

Given : The expression

$\displaystyle \sf{ {sin}^{2} \theta + \frac{1}{(1 + {tan}^{2} \theta) } }$

To find : The value of the expression

Formula :

$\displaystyle \sf{1. \: \: {sin}^{2} \theta + {cos}^{2} \theta } = 1$

$\displaystyle \sf{2. \: \: 1 + {tan}^{2} \theta = {sec}^{2} \theta }$

Solution :

Step 1 of 2 :

Write down the given expression

The given expression is

$\displaystyle \sf{ {sin}^{2} \theta + \frac{1}{(1 + {tan}^{2} \theta) } }$

Step 2 of 2 :

Find the value of the expression

$\displaystyle \sf{ {sin}^{2} \theta + \frac{1}{(1 + {tan}^{2} \theta) } }$

$\displaystyle \sf{ = {sin}^{2} \theta + \frac{1}{ {sec}^{2} \theta } } \: \: \: ( \: \because \: \: \: \: 1 + {tan}^{2} \theta = {sec}^{2} \theta \: )$