Question

Prove that: $\frac{Q^{2} - 1}{Q^2 + 1} + cosA = 0$ 
If $Q = cosA - cotA$

Answer 1

Δ Given,
                Q = cosecA - cotA
on squaring both side ,
 $Q^{2} = (cosecA-cotA)^{2} \\ = ( \frac{1-cosA}{sinA}) ^{2} \\ Q^{2} = \frac{ (1-cosA)^{2} }{(1- cos^{2}A) } \\ Q^{2} = \frac{(1-cosA)(1-cosA)}{(1-cosA)(1+cosA)} \\ Q^{2} = \frac{1-cosA}{1+cosA}$ 
$Q^{2} .(1+cosA)=1-cosA \\ Q^{2} + Q^{2} .cosA-1+cosA=0 \\ Q^{2} -1+ cosA(Q^{2} +1)=0 \\ \frac{ Q^{2}-1 }{ Q^{2}+1 } +cosA =0$