Question

sin2A/secA-1×sec2A/sec2A-1​

Answer 1

we have to find the value of $\left(\frac{sin2A}{secA-1}\right)\left(\frac{sec2A}{sec2A+1}\right)$

solution :

$\quad\left(\frac{sin2A}{secA-1}\right)\left(\frac{sec2A}{sec2A+1}\right)$

we know, sin2A = 2sinA cosA

sec2A = 1/cos2A

= $\left(\frac{2sinAcosA}{\frac{1}{cosA}-1}\right)\left(\frac{\frac{1}{cos2A}}{\frac{1}{cos2A}+1}\right)$

= $\frac{2sinAcos^2A}{1-cosA}\times\frac{1}{1+cos2A}$

using cos2A = 2cos²A -1

= $\frac{2sinAcos^2A}{1-cosA}\times\frac{1}{1+2cos^2A-1}$

= $\frac{sinA}{(1-cosA)}$

now using , sinA = 2sin(A/2)cos(A/2)

1 - cosA = 2sin²(A/2)

= $\frac{2sin(A/2)cos(A/2)}{2sin^2(A/2)}$

=$cot(A/2)$

Therefore $\left(\frac{sin2A}{secA-1}\right)\left(\frac{sec2A}{sec2A+1}\right)$ - cot(A/2).