Question

if sin theta + cos theta is equal to under root cos theta if sin theta + cos theta is equal to under root 2 cos theta theta is not 90 degree then the value of tan theta is under root 2 -1

$\sin(?) + \cos(?) = \sqrt{2} \cos(?)$
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Answer 1

Step-by-step explanation:

1. sin theta + cos theta = √2cos theta_____(1)

sin theta + cos theta/ cos theta= √2 cos theta/ cos theta

sin theta/ cos theta +1 = √2

tan theta +1 = √2

or

tan theta = √2-1

Answer 2

Correct Question :

if $\sin( \theta) + \cos( \theta)=\sqrt{2} \cos( \theta)$ and, $[(\theta) \ne0]$. Then show that $\tan(\theta)= \sqrt{2} - 1$

AnswEr :

$\longrightarrow\sin( \theta) + \cos( \theta) = \sqrt{2} \cos( \theta)$

$\longrightarrow(\sin( \theta) + \cos( \theta))^{2} = (\sqrt{2} \cos( \theta))^{2}$

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

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

$\longrightarrow1+ 2 \sin(\theta) \cos(\theta) = 2\cos^{2} ( \theta)$

$\longrightarrow2 \sin(\theta) \cos(\theta) = 2\cos^{2} ( \theta) - 1$

$\longrightarrow \sin2(\theta) = \cos2( \theta)$

$\longrightarrow \dfrac{\sin2(\theta)}{\cos2( \theta)} = 1$

$\longrightarrow \tan2( \theta) = 1$

$\longrightarrow \dfrac{2 \tan( \theta) }{1 - \tan^{2} (\theta)} = 1$

$\longrightarrow 2 \tan( \theta) = 1 - \tan^{2} (\theta)$

$\longrightarrow \tan^{2} (\theta) + 2 \tan( \theta) - 1 = 0$

$\longrightarrow\sf\tan(\theta) = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$\longrightarrow\sf\tan(\theta) = \dfrac{ - 2 \pm \sqrt{ {(2)}^{2} - 4 \times 1\times - 1} }{2 \times 1}$

$\longrightarrow\sf\tan(\theta) = \dfrac{ - 2 \pm \sqrt{4 + 4}}{2}$

$\longrightarrow\sf\tan(\theta) = \dfrac{ - 2 \pm \sqrt{8}}{2}$

$\longrightarrow\sf\tan(\theta) = \dfrac{ - 2 \pm 2\sqrt{2}}{2}$

$\longrightarrow\sf\tan(\theta) = \dfrac{ \cancel2( - 1 \pm \sqrt{2})}{\cancel2}$

$\longrightarrow\sf\tan(\theta) = - 1 \pm \sqrt{2}$

$\longrightarrow \boxed{\sf\tan(\theta) = \sqrt{2} - 1}$

⠀

$\boxed{\begin{minipage}{7 cm}\underline{\text{Trigonometric Identities Used Here}}\\ \\ \sin^2\theta + \cos^2\theta=1\\ \\2\sin\theta\cos\theta=\sin2\theta\\ \\2cos^2\theta - 1=\cos2\theta\\ \\ \tan2\theta=\dfrac{2\tan\theta}{1 - \tan^2\theta}\end{minipage}}$