Question

(1+tan^2 A )+(1+1/tan^2 A)=1/sin^2-sin^4A

Answer 1

Answer:

1/sin^2-sin^4A

Step-by-step explanation;

(1=tan^2a)+(1/tan^2a)=1/sin^24a

(1+tan^2a)(1+cot^2a)

sec^2a*cosec^2a

1/cos^2a*1/sin^2a

1/cos^2a*sin^2a

1/1-sin^2a*sin^2a=1/sin^2a-sin^4a

Answer 2

Answer:

Step-by-step explanation:

LHS:- (1+tan^2 A)+(1+1/tan^2 A)

= (sec^2 A)+(1+1/tan^2 A)                                   1+TAN^2 A = SEC^2 A

= (1/cos^2 A)+(tan^2 A+1/tan^2 A)   L.C.M - TAN^2 A   SEC^2 A = 1/COS^2 A

= (1/cos^2 A)+(sec^2 A/tan^2 A)

= (1/cos^2 A)+(cos^2 A / cos^2 A* sin^A)

= (1/cos^2 A)+(1/sin^2 A)                   L.C.M - sin^2 A * cos^2 A

= (sin^2 A/ sin^2 A * cos^2 A) + (cos^2 A/ sin^2 A * cos^2 A)

= (sin^2 A + cos^2 A) / (sin^2 A * cos^2 A)          SIN^2 A+COS^2 A = 1

= 1/sin^2 A * cos^2 A

= 1/sin2 A * (1-sin^2 A)                                         COS^2 A = 1 - SIN^2 A

= 1/sin^2 A - sin^4 A

RHS:- 1/sin^2 A - sin^4 A

RHS=LHS

HENCE PROVED....

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