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Question: If \(2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}\), then \(x = \frac{5\pi}{4}\) equals...

If 2sin⁡x+2cos⁡x>21−(1/2)2^{\sin x} + 2^{\cos x} > 2^{1 - (1/\sqrt{2})}, then x=5π4x = \frac{5\pi}{4} equals

A

(1+tan⁡θ)(1+tan⁡φ)=2⇒tan⁡θ+tan⁡φ1−tan⁡θtan⁡φ=1(1 + \tan\theta)(1 + \tan\varphi) = 2 \Rightarrow \frac{\tan\theta + \tan\varphi}{1 - \tan\theta\tan\varphi} = 1

B

⇒\Rightarrow

C

tan⁡(θ+φ)=1\tan(\theta + \varphi) = 1

D

⇒\Rightarrow

Answer

tan⁡(θ+φ)=1\tan(\theta + \varphi) = 1

Explanation

Solution

1sin⁡2C2−1sin⁡2B2=1sin⁡2B2−1sin⁡2A2\frac{1}{\sin^{2}\frac{C}{2}} - \frac{1}{\sin^{2}\frac{B}{2}} = \frac{1}{\sin^{2}\frac{B}{2}} - \frac{1}{\sin^{2}\frac{A}{2}}

ab(s−a)(s−b)−ac(s−a)(s−c)\frac{ab}{(s - a)(s - b)} - \frac{ac}{(s - a)(s - c)} ⇒=ac(s−a)(s−c)−bc(s−b)(s−c)= \frac{ac}{(s - a)(s - c)} - \frac{bc}{(s - b)(s - c)}.