Question

1/ sec²1 by sin square theta minus cos square theta + 1 by cosec squared theta minus sin square theta the whole into sin square theta cos square theta is equal to 1 minus sin square theta cos square theta by 2 + sin square theta cos square theta ​

Answer 1

Step-by-step explanation:

(1/Sec^2 theta - Cos^2 theta} + (1/Cosec^2 theta - Sin^2 theta ) x Sin^2 theta Cos^2 theta

= (1 -Sin^2 theta Cos^2 theta)(2 + Sin^2 theta Cos^2 theta}

LHS = (1/(Sec^2 theta - Cos^2 theta) + (1 /Cosec^2 theta - Sin^2 theta)  Sin^2 theta x Cos^2 theta  

Secθ = 1/Cosθ  & Cosecθ = 1/Sinθ

= [1/(1/Cos^2 theta) - Cos^2 theta] + [1/(1/Sin^2 theta)] - Sin^2 theta] x Sin^2\theta Cos^2\theta  

= ({Cos^2 theta}{1 - Cos^4 theta} +(Sin^2 theta}{1 - Sin^4 theta}) x Sin^2 theta Cos^2 theta  

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

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

= ({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)})  Sin^2\theta Cos^2\theta

={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}

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

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

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

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

= RHS

HOPE THIS HELPS..............

Answer 2

Answer:

Step-by-step explanation:

Lhs= 1 upon sec²theta - cos²theta + 1 upon cosec²theta -sin² theta×sin²theta × cos² theta