Question

$cosec\theta (sec\theta -1)-cot\theta (1-cos\theta )=tan\theta -sin\theta$

Answer 1

$LHS=cosec\theta (sec\theta -1)-cot\theta (1-cos\theta )$

$=\frac{1}{sin\theta }(\frac{1}{cos\theta }-1)-\frac{cos\theta }{sin\theta }(1-cos\theta )[\because cosec\theta =\frac{1}{sin\theta },sec\theta =\frac{1}{cos\theta }cot\theta =\frac{cos\theta }{sin\theta }]$

$=\frac{(1-cos\theta )}{sin\theta cos\theta }-\frac{cos\theta (1-cos\theta )}{sin\theta }=\frac{(1-cos\theta )-co{s}^2\theta (1-cos\theta )}{sin\theta cos\theta }$

$=\frac{(1-cos\theta )(1-co{s}^2\theta )}{sin\theta cos\theta }=\frac{(1-cos\theta )si{n}^2\theta }{sin\theta cos\theta }(\because 1-co{s}^2\theta =si{n}^2\theta )$

$=(1-cos\theta )\frac{sin\theta }{cos\theta }$

$=\frac{sin\theta }{cos\theta }-sin\theta =tan\theta -sin\theta (\because tan\theta =sin\theta -cos\theta )$

=RHS

Hence proved.