Question

Find the value of
$\sf \: sin^{2}O +\frac{1}{1+tan^{2}O}$
​

Answer 1

Answer:-

$\red{\bigstar}$ Value of $\bf sin^2 \theta + \dfrac{1}{1+tan^2 \theta} \large\leadsto{\tt\purple{1}}$

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• Given:-

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

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• Solution:-

➪ $\sf sin^2 \theta + \dfrac{1}{1+tan^2 \theta}$

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We know,

$\purple{\bigstar}$ $\underline{\boxed{\bf\green{sec^2 \theta = 1+tan^2 \theta}}}$

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➪ $\sf sin^2 \theta + \dfrac{1}{sec^2 \theta}$

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Also,

$\purple{\bigstar}$ $\underline{\boxed{\bf\green{cos \theta = \dfrac{1}{sec \theta}}}}$

➪ $\sf sin^2 \theta + cos^2 \theta$

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We know,

$\purple{\bigstar}$ $\underline{\boxed{\bf\green{sin^2 \theta + cos^2 \theta = 1}}}$

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★ $\large{\bf\pink{1}}$

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Therefore, the value of $\bf sin^2 \theta + \dfrac{1}{1+tan^2 \theta}$ is 1.

Answer 2

Answer:

$\sf{➙\:sin^{2}\theta +\dfrac{1}{1+tan^{2}\theta } \leadsto}$ $\boxed{\bold{\red{1}}}$

$\:$

To find :-

$\sf{➪\:sin^{2}\theta +\dfrac{1}{1+tan^{2}\theta }}$

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Solution :-

$\sf{➪\:sin^{2}\theta +\dfrac{1}{1+tan^{2}\theta }}$

$\sf{= sin^{2}\theta +\dfrac{1}{sec²\theta}}$

$\sf{=sin^{2}\theta +cos²\theta=}$$\bold{\red{1}}$

$\:$

Formula to remember:

$\sf{•\:sin²\theta + cos²\theta = 1}$

$\sf{•\:sec²\theta - tan²\theta = 1}$

$\sf{•\:cosec²\theta + cot²\theta = 1}$