Question

if sec^4 theta + sec^2 theta = 10+ tan^4 theta+tan^2 theta then sin^2 theta =

Answer 1

ANSWER:

Given:

• sec⁴θ + sec²θ = 10 + tan⁴θ + tan²θ

To Find:

• sin²θ

Solution:

We are given that,

$\implies\sec^4\theta+\sec^2\theta=10+\tan^4\theta+\tan^2\theta$

Taking sec²θ common in LHS,

$\implies\sec^2\theta(\sec^2\theta+1)=10+\tan^4\theta+\tan^2\theta$

Taking tan²θ common in RHS,

$\implies\sec^2\theta(\sec^2\theta+1)=10+\tan^2\theta(\tan^2\theta+1)$

We know that,

$\hookrightarrow\sec^2\phi=1+\tan^2\phi$

So,

$\implies\sec^2\theta(\sec^2\theta+1)=10+\tan^2\theta(\tan^2\theta+1)$

$\implies\sec^2\theta(\sec^2\theta+1)=10+\tan^2\theta(\sec^2\theta)$

$\implies\sec^2\theta(\sec^2\theta+1)=10+\tan^2\theta\sec^2\theta$

Transposing tan²θ sec²θ to LHS,

$\implies\sec^2\theta(\sec^2\theta+1)-\tan^2\theta\sec^2\theta=10$

Taking sec²θ common,

$\implies\sec^2\theta(\sec^2\theta+1-\tan^2\theta)=10$

$\implies\sec^2\theta(\sec^2\theta-\tan^2\theta+1)=10$

We know that,

$\hookrightarrow\sec^2\phi-\tan^2\phi=1$

So,

$\implies\sec^2\theta(1+1)=10$

$\implies2\sec^2\theta=10$

So,

$\implies\sec^2\theta=5$

We know that,

$\hookrightarrow\sec\phi=\dfrac{1}{\cos\phi}$

So,

$\implies\sec^2\theta=5$

$\implies\dfrac{1}{\cos^2\theta}=5$

So,

$\implies\cos^2\theta=\dfrac{1}{5}$

We know that,

$\hookrightarrow\sin^2\phi=1-\cos^2\phi$

So,

$\implies\cos^2\theta=\dfrac{1}{5}$

$\implies1-\sin^2\theta=\dfrac{1}{5}$

$\implies\sin^2\theta=1-\dfrac{1}{5}$

$\implies\sin^2\theta=\dfrac{5-1}{5}$

Hence,

$\implies\bf sin^2\theta=\dfrac{4}{5}$

Therefore, the value of sin²θ is ⅘.

Answer 2

Answer:

4/5

Step-by-step explanation:

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