Math, asked by Anonymous, 1 year ago

solve this question please

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kvnmurty: is it correct? ?
Anonymous: yeah
Anonymous: send me the answer
kvnmurty: no.. put theta = 90... LHS =2. RHS =1... Question incorrect
Anonymous: I think so u can use this method in LHS as (a-b)²

Answers

Answered by kvnmurty
0
There is a mistake in the given problem... Exponent 2 is missing in the denominator on the LHS.

See the solution.
X=\frac{1-cos\theta+sin\theta}{1+cos\theta+sin\theta}=\frac{2Sin^2 \frac{\theta}{2}+2 sin \frac{\theta}{2} Cos \frac{\theta}{2}}{2 Cos^2 \frac{\theta}{2}+2 sin \frac{\theta}{2} Cos \frac{\theta}{2}}\\\\=\frac{Sin \frac{\theta}{2}}{Cos \frac{\theta}{2}}\\\\\frac{1-cos\theta}{1+cos\theta}=\frac{2Sin^2 \frac{\theta}{2}}{2Cos^2 \frac{\theta}{2}}\\\\So \: \ X^2=LHS=RHS

kvnmurty: :-) :-)
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