Question

(1+tan A. tanB)²+
(tanA-tan B)²=sec²A.sec²B​

Answer 1

Step-by-step explanation:

1+tanA^2.tanB^2+2tanA.tanB+tanA^2+tanB^2

-2tanA.tanB

=1+tanA^2(1+tanB^2)+tanB^2

=1+tanA^2.secB^2+tanB^2

=1+tanA^2/cosB^2+sinB^2/cosB^2

=cosB^2+(tanA^2+sinB^2)/cosB^2

=(sinB^2+cosB^2)+tanA^2/cosB^2

=1+tanA^2/cosB^2

=secA^2.secB^2(proved)

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Answer 2

Solution:-

(1 + tan A . tanB)²+ (tanA - tan B)² = sec²A . sec²B

Now,

(1 + tanA . tanB)²+ (tanA - tanB)² = LHS

= (1² + 2tanA . tanB + tan²A . tan²B) + (tan²A - 2tanA . tanB + tan²B)

= 1 + 2tanA . tanB + tan²A.tan²B + tan²A - 2tanA . tanB + tan²B

= 1 + tan²A . tan²B + tan²A + tan²B

= 1 + tan²B( tan²A + 1)

= 1 + tan²B . (tan²A + 1)

= sec²B . sec²A = sec²A . sec²B

But sec²A . sec²B = RHS

•°• LHS = RHS

•°•(1 + tanA . tanB)² + (tanA - tanB)² = sec²A . sec²B

Hence proved that,

Hence proved that,(1 + tanA . tanB)² + (tanA - tanB)² = sec²A . sec²B