Math, asked by AtharvNimbalkar, 1 year ago

PROVE THAT
1 + cos 0/
sin 0 -
sin0/1 + cos0 =
2 cot0


sin
1 + cos
2 cote.​

Answers

Answered by Anonymous
8

Step-by-step explanation:

Given,</p><p></p><p>Prove:  \frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}=\frac{1}{sec\theta-tan\theta}</p><p></p><p>LHS:</p><p></p><p>\frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}</p><p></p><p>=\frac{\frac{sin\theta-cos\theta+1}{cos\theta} }{\frac{sin\theta+cos\theta-1}{cos\theta} } (By dividing cos\theta on numerator and denominator)</p><p></p><p>=\frac{tan\theta-1+sec\theta}{tan\theta+1-sec\theta}</p><p></p><p>=\frac{(tan\theta+sec\theta)-(sec^{2}\theta-tan^{2}\theta)  }{1-sec\theta+tan\theta}</p><p></p><p>=\frac{(tan\theta+sec\theta)-(sec\theta-tan\theta)(sec\theta+tan\theta)  }{1-sec\theta+tan\theta}</p><p></p><p>=\frac{(tan\theta+sec\theta)(1-sec\theta+tan\theta)}{1-sec\theta+tan\theta}</p><p></p><p>=tan\theta+sec\theta</p><p></p><p>=\frac{(sec\theta+tan\theta)(sec\theta-tan\theta)}{(sec\theta-tan\theta)}</p><p></p><p>=\frac{(sec^{2} \theta-tan^{2} \theta)}{(sec\theta-tan\theta)}</p><p></p><p>=\frac{1}{sec\theta-tan\theta}=RHS</p><p></p><p>Hence Proved.

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