Question

One upon sec squared theta minus cos square theta + 1 upon cosec squared theta minus sin square theta x squared theta into cos square theta is equal to 1 minus sin square theta into cos square theta upon 2 + sin square theta cos theta

Answer 1

Answer:

Proved

Step-by-step explanation:

$(\frac{1}{Sec^2\theta - Cos^2\theta} + \frac{1}{Cosec^2\theta - Sin^2\theta} )\times Sin^2\theta Cos^2\theta = \frac{1 -Sin^2\theta Cos^2\theta}{2 + Sin^2\theta Cos^2\theta}$

LHS = $(\frac{1}{Sec^2\theta - Cos^2\theta} + \frac{1}{Cosec^2\theta - Sin^2\theta}) \times Sin^2\theta Cos^2\theta$

Secθ = 1/Cosθ  & Cosecθ = 1/Sinθ

= $(\frac{1}{\frac{1}{Cos^2\theta} - Cos^2\theta} + \frac{1}{\frac{1}{Sin^2\theta} - Sin^2\theta}) \times Sin^2\theta Cos^2\theta$

= $(\frac{Cos^2\theta}{1 - Cos^4\theta} + \frac{Sin^2\theta}{1 - Sin^4\theta}) \times Sin^2\theta Cos^2\theta$

= $(\frac{Cos^2\theta}{(1 + Cos^2\theta)(1 - Cos^2\theta)} + \frac{Sin^2\theta}{(1 + Sin^2\theta)(1 - Sin^2\theta)}) \times Sin^2\theta Cos^2\theta$

= $(\frac{Cos^2\theta}{(1 + Cos^2\theta)(Sin^2\theta)} + \frac{Sin^2\theta}{(1 + Sin^2\theta)(Cos^2\theta)}) \times Sin^2\theta Cos^2\theta$

= $(\frac{Cos^4\theta + Cos^4\theta Sin^2\theta + Sin^4\theta + Sin^4\theta Cos^2\theta}{(1 + Cos^2\theta)(Sin^2\theta)(1 + Sin^2\theta)(Cos^2\theta)}) \times Sin^2\theta Cos^2\theta$

= $\frac{Cos^4\theta + Cos^2\theta Sin^2\theta (Cos^2\theta + Sin^2\theta) + Sin^4\theta }{1 + Cos^2\theta + Sin^2\theta + Cos^2\theta Sin^2\theta}$

= $\frac{Cos^4\theta + Cos^2\theta Sin^2\theta (1) + Sin^4\theta }{1 + 1 + Cos^2\theta Sin^2\theta}$

= $\frac{ (Cos^2\theta + Sin^2\theta )^2 - Cos^2\theta Sin^2\theta}{2+ Cos^2\theta Sin^2\theta}$

= $\frac{ (1 )^2 - Cos^2\theta Sin^2\theta}{2+ Cos^2\theta Sin^2\theta}$

= $\frac{ 1 - Sin^2\theta Cos^2\theta }{2+ Sin^2\theta Cos^2\theta}$

= RHS

QED

Answer 2

Answer:

Step-by-step explanation: