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Question: \(\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}\)=...

1+sin⁡A−cos⁡A1+sin⁡A+cos⁡A\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A}=

A

sin⁡A2\sin\frac{A}{2}

B

cos⁡A2\cos\frac{A}{2}

C

tan⁡A2\tan\frac{A}{2}

D

cot⁡A2\cot\frac{A}{2}

Answer

tan⁡A2\tan\frac{A}{2}

Explanation

Solution

1+sin⁡A−cos⁡A1+sin⁡A+cos⁡A\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} =2sin⁡2A2+2sin⁡A2cos⁡A22cos⁡2A2+2sin⁡A2cos⁡A2= \frac{2\sin^{2}\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}{2\cos^{2}\frac{A}{2} + 2\sin\frac{A}{2}\cos\frac{A}{2}}

=2sin⁡A2(sin⁡A2+cos⁡A2)2cos⁡A2(cos⁡A2+sin⁡A2)= \frac{2\sin\frac{A}{2}\left( \sin\frac{A}{2} + \cos\frac{A}{2} \right)}{2\cos\frac{A}{2}\left( \cos\frac{A}{2} + \sin\frac{A}{2} \right)}=tan⁡A2\tan\frac{A}{2}.

Trick : Put A=60o.A = 60^{o}.

Then 1+(3/2)−(1/2)1+(3/2)+(1/2)=1+33+3=13\frac{1 + (\sqrt{3}/2) - (1/2)}{1 + (\sqrt{3}/2) + (1/2)} = \frac{1 + \sqrt{3}}{3 + \sqrt{3}} = \frac{1}{\sqrt{3}}

which is given by option (3), i.e. tan⁡60o2=13\tan\frac{60^{o}}{2} = \frac{1}{\sqrt{3}}

Note : Students should remember at the time of assuming the values of A, B, θ, ..... etc. that, for the assumed values, the options must have different values.