Question

Prove sin squared theta + 1 upon 1 + 10 square theta is equals to 1

Answer 1

Answer:

1

Step-by-step explanation:

${ \sin ^{2} ( \alpha ) } + \frac{1}{1 + \tan ^{2} \alpha \ } = 1 \\ { \sin {}^{2} \alpha } + \frac{ \cos {}^{2} \alpha }{ \cos {}^{2} \alpha + \sin {}^{2} \alpha } = 1 \\ \\ \sin {}^{2} \alpha + \cos {}^{2} \alpha = 1 \\ 1 = 1 \\ lhs \: = rhs \: \\ \: proved$

Answer 2

$sin^2 \theta + \frac{1}{1+ tan^2 \theta} = 1$

Solution:

Given that, we have to prove,

$sin^2 \theta + \frac{1}{1+ tan^2 \theta} = 1$

Take the L.H.S

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

We know that,

$1+tan^2 \theta = sec^2 \theta$

Therefore,

$\frac{1}{1+tan^2 \theta} = \frac{1}{sec^2 \theta}$

Thus the LHS becomes,

$sin^2 \theta + \frac{1}{sec^2 \theta}$

We know that,

$cos^2 \theta = \frac{1}{sec^2 \theta}$

Therefore, the above equation becomes

$sin^2 \theta + cos^2 \theta$

By trignometric identity,

$sin^2 \theta + cos^2 \theta = 1$

Thus, LHS = 1

Therefore,

$sin^2 \theta + \frac{1}{1+ tan^2 \theta} = 1$

Thus proved

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