Question

If cos theetha +cos2 theetha =1 then sin 2 theetha+sin4thhetha =

Answer 1

✌️Hey dude here is your answer________✌️⬆️

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• Given :-

sinθ + sin²θ = 1

• To find :-

The value of : cos²θ + cos⁴θ

• Salutation :-

sinθ + sin²θ = 1

sinθ = 1 - sin²θ

sinθ = cos²θ ---------- ( i )

[ • As sin²θ + cos²θ = 1

So , sin²θ = 1 - cos²θ ]

★ Method - 1

sinθ = cos²θ

( sinθ )² = ( cos²θ )²

sin²θ = cos⁴θ

1 - cos²θ = cos⁴θ

cos⁴θ + cos²θ = 1 [ ★ Required answer ]

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★ Method - 2

cos²θ + cos⁴θ

= sinθ + ( sinθ )²

[ • Putting the value of cos²θ = sinθ ]

= sinθ + sin²θ

= 1 [ • Given , sinθ + sin²θ = 1 ]

• So finally ,

[ cos²θ + cos⁴θ = 1 ]

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★ Be Brainly

Answer 2

AnswEr:

$\underline{\bigstar\:\textsf{According \: to \: given \: in \: question:}}$

$\normalsize\underline\textsf{Left \: hand \: side:}$

$\normalsize\ : \implies\sf\ cos\theta + cos^2\theta = 1 \\ \\ \normalsize\ : \implies\sf\ cos = 1 - cos^2\theta$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{sin^2\theta = 1 - cos^2\theta }) }$

$\normalsize\ : \implies\sf\ cos\theta = sin^2\theta \: \: \: ---(eq.1)$

$\rule{100}2$

$\normalsize\underline\textsf{Now; \: Right \: hand \: side:}$

$\normalsize\ : \implies\sf\ sin^2\theta + sin^4\theta \\ \\ \normalsize\ : \implies\sf\ sin^2\theta + (sin^2\theta)^2$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \red{Putting \: values \: from \: equation \: 1}) }$

$\normalsize\ : \implies\sf\ cos\theta + (cos\theta)^2 \\ \\ \normalsize\ : \implies\sf\ cos\theta + cos^2\theta \\ \\ \normalsize\ : \implies\sf\ 1 \\ \\ \normalsize\ : \implies\sf\ sin^2\theta + sin^4\theta = 1$

$\normalsize\ : \implies{\underline{\boxed{\sf \purple{sin^2\theta + sin^4\theta = 1}}}}$

$\rule{100}2$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\cos^2\theta=1-\sin^2\theta \\ \\ 4)1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5) \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\sec^2\theta=1+\tan^2\theta \\ \\ 8)\sec^2\theta-\tan^2\thetha=1 \\ \\ 9)\tan^2\theta=\sec^2\theta-1 \end{minipage}}$