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Question: If \(\sin\theta + \sin\varphi = a\)and \(\cos\theta + \cos\varphi = b,\) then \(\tan\frac{\theta - \...

If sin⁡θ+sin⁡φ=a\sin\theta + \sin\varphi = aand cos⁡θ+cos⁡φ=b,\cos\theta + \cos\varphi = b, then tan⁡θ−φ2\tan\frac{\theta - \varphi}{2}is equal to

A

a2+b24−a2−b2\sqrt{\frac{a^{2} + b^{2}}{4 - a^{2} - b^{2}}}

B

4−a2−b2a2+b2\sqrt{\frac{4 - a^{2} - b^{2}}{a^{2} + b^{2}}}

C

a2+b24+a2+b2\sqrt{\frac{a^{2} + b^{2}}{4 + a^{2} + b^{2}}}

D

4+a2+b2a2+b2\sqrt{\frac{4 + a^{2} + b^{2}}{a^{2} + b^{2}}}

Answer

4−a2−b2a2+b2\sqrt{\frac{4 - a^{2} - b^{2}}{a^{2} + b^{2}}}

Explanation

Solution

Given that sin⁡θ+sin⁡φ=a\sin\theta + \sin\varphi = a …..(i)

and cos⁡θ+cos⁡φ=b\cos\theta + \cos\varphi = b …..(ii)

Squaring, sin⁡2θ+sin⁡2φ+2sin⁡θsin⁡φ=a2\sin^{2}\theta + \sin^{2}\varphi + 2\sin\theta\sin\varphi = a^{2}

and cos⁡2θ+cos⁡2φ+2cos⁡θcos⁡φ=b2\cos^{2}\theta + \cos^{2}\varphi + 2\cos\theta\cos\varphi = b^{2}

Adding, 2+ 2 (sin⁡θsin⁡φ+cos⁡θcos⁡φ)=a2+b2(\sin\theta\sin\varphi + \cos\theta\cos\varphi) = a^{2} + b^{2}

⇒2cos⁡(θ−φ)=a2+b2−22\cos(\theta - \varphi) = a^{2} + b^{2} - 2⇒ cos⁡(θ−φ)=a2+b2−22\cos(\theta - \varphi) = \frac{a^{2} + b^{2} - 2}{2}

⇒1−tan⁡2θ−φ21+tan⁡2θ−φ2=a2+b2−22\Rightarrow \frac{1 - \tan^{2}\frac{\theta - \varphi}{2}}{1 + \tan^{2}\frac{\theta - \varphi}{2}} = \frac{a^{2} + b^{2} - 2}{2}

⇒ (a2+b2)+(a2+b2)tan⁡2θ−φ2−2−2tan⁡2θ−φ2(a^{2} + b^{2}) + (a^{2} + b^{2})\tan^{2}\frac{\theta - \varphi}{2} - 2 - 2\tan^{2}\frac{\theta - \varphi}{2}

=2−2tan⁡2θ−φ2= 2 - 2\tan^{2}\frac{\theta - \varphi}{2} ⇒4−a2−b2a2+b2=tan⁡2θ−φ2\frac{4 - a^{2} - b^{2}}{a^{2} + b^{2}} = \tan^{2}\frac{\theta - \varphi}{2}

⇒ tan⁡(θ−φ)2=4−a2−b2a2+b2\tan\frac{(\theta - \varphi)}{2} = \sqrt{\frac{4 - a^{2} - b^{2}}{a^{2} + b^{2}}}

Trick : Put θ=π2,φ=0o\theta = \frac{\pi}{2},\varphi = 0^{o}, then a=1=ba = 1 = b

∴tan⁡θ−φ2=1\tan\frac{\theta - \varphi}{2} = 1, which is given by (1) and (2).

Again putting θ=π4=φ\theta = \frac{\pi}{4} = \varphi, we get tan⁡θ−φ2=0\tan\frac{\theta - \varphi}{2} = 0, which is given by (2).