Question

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Answer 1

Given : Sinθ + Sin²θ = 1

To Find : Cos²θ + Cos⁴θ =  

A  - 1

B     1

C    0

D   None of these

Solution:

Sinθ + Sin²θ = 1

=> Sinθ =  1- Sin²θ

=>  Sinθ = Cos²θ

Squaring both sides

=> Sin²θ = Cos⁴θ

Cos²θ + Cos⁴θ

Cos²θ  = Sinθ    and  Cos⁴θ  = Sin²θ

= Sinθ + Sin²θ

=  1

Cos²θ + Cos⁴θ  = 1

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If sinθ + cosθ = √2cosθ, (θ ≠ 90°) then the value of tanθ is a) √2 ...

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Answer 2

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

$\sf{ \sin \theta \: = \: 1 - { \sin}^{2} \theta}$

$\boxed{ \boxed{ \sin \theta \: = \: { \cos}^{2} \theta}} - - - (1)$

$\sf \red{from \: (1)}$

$\sin \theta \: = \: { \cos}^{2} \theta$

$\sf\red{S.Q.B.S \: we \: get : }$

${ \sin}^{2} \theta \: = { ({ \cos}^{2} )}^{2} \theta$

$\boxed{\boxed{ { \sin}^{2} \theta \: = { \cos }^{4} \theta}} - - - (2)$

$\sf \red{ By \: adding\: (1) \& (2) \: we \: get : }$

$\sin \theta \: + \: { \sin}^{2} \theta \: = {\cos}^{2} \theta \: + { \cos}^{4} \theta \: = 1$

$\sf \red{ Answer: }$

$\boxed{ \boxed{ {\cos}^{2} \theta \: + { \cos}^{4} \theta \: = 1}}$