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Question: If \(\theta\)and \(\varphi\)are angles in the 1st quadrant such that \(\tan\theta = 1/7\)and \(\sin\...

If θ\thetaand φ\varphiare angles in the 1st quadrant such that tan⁡θ=1/7\tan\theta = 1/7and sin⁡φ=1/10\sin\varphi = 1/\sqrt{10}.Then

A

θ+2φ=90∘\theta + 2\varphi = 90{^\circ}

B

θ+2φ=60∘\theta + 2\varphi = 60{^\circ}

C

θ+2φ=30∘\theta + 2\varphi = 30{^\circ}

D

θ+2φ=45∘\theta + 2\varphi = 45{^\circ}

Answer

θ+2φ=45∘\theta + 2\varphi = 45{^\circ}

Explanation

Solution

Given, tan⁡θ=17,sin⁡φ=110\tan\theta = \frac{1}{7},\sin\varphi = \frac{1}{\sqrt{10}}

sin⁡θ=150,cos⁡θ=750,cos⁡φ=310\sin\theta = \frac{1}{\sqrt{50}},\cos\theta = \frac{7}{\sqrt{50}},\cos\varphi = \frac{3}{\sqrt{10}}

∴cos⁡2φ=2cos⁡2φ−1=2.910−1=810\therefore\cos 2\varphi = 2\cos^{2}\varphi - 1 = 2.\frac{9}{10} - 1 = \frac{8}{10}

sin⁡2φ=2sin⁡φcos⁡φ=2×.110×310=610\sin 2\varphi = 2\sin\varphi\cos\varphi = 2 \times .\frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \frac{6}{10}

∴ cos⁡(θ+2φ)=cos⁡θcos⁡2φ−sin⁡θsin⁡2φ\cos(\theta + 2\varphi) = \cos\theta\cos 2\varphi - \sin\theta\sin 2\varphi

=750×810−150.610= \frac{7}{\sqrt{50}} \times \frac{8}{10} - \frac{1}{\sqrt{50}}.\frac{6}{10}

=56−61050=501050=5210=12= \frac{56 - 6}{10\sqrt{50}} = \frac{50}{10\sqrt{50}} = \frac{5\sqrt{2}}{10} = \frac{1}{\sqrt{2}}

∴θ+2φ=45o\theta + 2\varphi = 45^{o}.