Question

If $a=sec\theta -tan\theta$ and $b=cosec\theta +cot\theta$, then show that ab+a-b+1=0.

Answer 1

LHS=ab+a-b+1

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

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

$=\frac{1}{sin\theta cos\theta }+\frac{1}{cos\theta }\times \frac{cos\theta }{sin\theta }-\frac{sin\theta }{cos\theta }\times \frac{1}{sin\theta }-tan\theta \times cot\theta +\frac{1}{cos\theta }-\frac{sin\theta }{cos\theta }-\frac{1}{sin\theta }-\frac{cos\theta }{sin\theta }+1$

$=\frac{1}{sin\theta cos\theta }+\frac{1}{sin\theta }-\frac{1}{cos\theta }-1+\frac{1}{cos\theta }-\frac{1}{sin\theta }-\frac{cos\theta }{sin\theta }+1=\frac{1}{sin\theta cos\theta }-\frac{sin\theta }{cos\theta }-\frac{cos\theta }{sin\theta }$

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

$=\frac{1-1}{sin\theta .cos\theta }=0=RHS$. Hence Proved.