Question

show that (1 + tan ^2) + ( 1 + 1/tan^2 A = 1/ sin ^2-sin ^2A​

Answer 1

Step-by-step explanation:

Given Qusetion :-

(1+ Tan^2 A) + ( 1+ 1/ Tan^2 A ) =1/ sin ^2-sin ^2A

Correct Question:-

Show that

(1+ Tan^2 A) + ( 1+ 1/ Tan^2 A ) =

1/Sin^2 A- Sin^4 A

Solution:-

LHS:-

(1+ Tan^2 A) + ( 1+ 1/ Tan^2 A )

=> (1+ Tan^2 A) + ( 1+ Cot^2 A)

Since Cot A = 1/ Tan A

We know that

Sec^2 A - Tan^2 A = 1

=> Sec^2 A = 1+ Tan^2 A

and

Cosec^2 A - Cot^2 A = 1

=> Cosec^2 A = 1+ Cot^2 A

now

(1+ Tan^2 A) + ( 1+ Cot^2 A)

=> Sec^2 A + Cosec^2

=> (1/Cos^2 A) + (1/ Sin^2 A)

Since ,Sec A= 1/ Cos A

Cosec A = 1/ Sin A

=> (Sin^2 A+ Cos^2 A)/ Sin^2 A Cos^2 A

=> 1/Sin^2 A Cos^2 A

We know that

Sin^2 A + Cos^2 A = 1

=> 1/Sin^2 A (1-Sin^2 A)

=> 1/ Sin^2 A - Sin^4 A

=> RHS

LHS = RHS

Used formulae:-

• Cot A = 1/ Tan A
• Sec^2 A - Tan^2 A = 1
• Sec A= 1/ Cos A
• Cosec A = 1/ Sin A
• Cosec^2 A - Cot^2 A = 1
• Sin^2 A + Cos^2 A = 1