Question

8
Prove that
cos0/1-tan0
+sin0/1-cot0 =
sin 0 + cos 0​

Answer 1

Answer:

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Answer 2

Answer:

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