Question

If cosθ+secθ=√2,find the value of cos²θ+sec²θ​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{cos\theta+sec\theta=\sqrt{2}}$

$\underline{\textsf{To find:}}$

$\textsf{The value of}$

$\mathsf{cos^2\theta+sec^2\theta}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{cos\theta+sec\theta=\sqrt{2}}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{(cos\theta+sec\theta)^2=(\sqrt{2})^2}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{(a+b)^2=a^2+b^2+2ab}}$

$\mathsf{cos^2\theta+sec^2\theta+2\;cos\theta\;sec\theta=2}$

$\mathsf{cos^2\theta+sec^2\theta+2\;cos\theta(\dfrac{1}{cos\theta})=2}$

$\mathsf{cos^2\theta+sec^2\theta+2=2}$

$\mathsf{cos^2\theta+sec^2\theta=2-2}$

$\implies\boxed{\mathsf{cos^2\theta+sec^2\theta=0}}$

Answer 2

SOLUTION

GIVEN

$\sf{ \cos \theta + \sec \theta = \sqrt{2} \: }$

TO DETERMINE

$\sf{ {\cos}^{2} \theta + { \sec}^{2} \theta \: }$

FORMULA TO BE IMPLEMENTED

We are aware of the identity that

$\sf{ {a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab\: }$

EVALUATION

$\sf{ {\cos}^{2} \theta + { \sec}^{2} \theta \: }$

$= \sf{ { ( \cos \theta + \sec \theta \:) }^{2} - 2 \times \cos \theta \times \sec \theta }$

$\displaystyle = \sf{ { ( \cos \theta + \sec \theta \:) }^{2} - 2 \times \cos \theta \times \frac{1}{\cos \theta} }$

$\displaystyle = \sf{ { ( \cos \theta + \sec \theta \:) }^{2} - 2 }$

$\sf{ = {( \sqrt{2} )}^{2} - 2\: }$

$= 2 - 2$

$= 0$

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