Question

tan²A - tan²B = (sin²A - sin²B)/(cos²A * cos²B) prove the question given above​

Answer 1

Answer:

Step-by-step explanation:

tan²x = sin²x/cos²x, and sin²x + cos²x = 1

tan²A - tan²B = sin²A/cos²A - sin²B/cos²B  

= sin²A cos²B/cos²A cos²B - sin²B cos²A/cos²A cos²B  

= sin²A(1-sin²B)/cos²A cos²B - sin²B(1-sin²A)/cos²A cos²B  

= (sin²A - sin²Asin²B)/cos²A cos²B - (sin²B - sin²Asin²B)/cos²A cos²B  

= (sin²A - sin²Asin²B) - (sin²B - sin²Asin²B)/cos²A cos²B  

= (sin²A - sin²Asin²B - sin²B + sin²Asin²B)/cos²A cos²B  

= (sin²A - sin²B)/cos²A cos²B

Answer 2

Step-by-step explanation:

Given:-

Tan^2 A - Tan^2 B

To find:-

Prove that :Tan^2 A - Tan^2 B

= (Sin^2 A - Sin^2 B )/(Cos^2 A Cos^2 B)

Solution:-

LHS:-

Tan^2 A - Tan^2 B

We know that

Tan θ = Sin θ / Cos θ

=> (Sin A / Cos A)^2 - (Sin B/ Cos B)^2

=> Sin^2 A / Cos^2 A - Sin^2 B / Cos^2 B

=> (Sin^2 A Cos^2 B - Cos^2 A Sin^2 B ) /

Cos^2 A Cos^2 B

We know that

Sin^2 A + Cos^2 A = 1

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

Cos^2 A Cos^2 B

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

=> (Sin^2 A - Sin^2 B )/(Cos^2 A Cos^2 B)

=> RHS

LHS=RHS

Hence, Proved

Answer:-

Tan^2 A - Tan^2 B

= (Sin^2 A - Sin^2 B )/(Cos^2 A Cos^2 B)

Used formulae:-

• Tan θ = Sin θ / Cos θ

• Sin^2 A + Cos^2 A = 1