prove the identity tan^2x+1/tan^2x=csc^2x?

Question

and this one too : If cos A = -7/25 and sin B = 4/5 with A in QIII and B in QII, find ht eexact value of the following expressions.
sin A, tan B, cos(A+B), sin2B, cos A/2

Answer 1

[tan^2(x)+1] / tan^2(x) = [tan^2(x) / tan^2(x)] + [1 / tan^2(x)] = 1 + cot^2(x) = csc^2(x).

Note that 1 + cot^2(x) = csc^2(x) is a standard identity that you can use directly.

For your other questions, you can use the given information about cos(A) and sin(B) to set up a right triangle where you know the lengths of two sides and can use the Pythagorean theorem to find the third. You can then apply the basic trigonometric functions in a straightforward fashion. The angle sum, double angle, and half angle trig identities are tabulated in my second reference, and are probably in your textbook as well.

Answer 2

tan^2x+1/tan^2x =1 +1/tan^2x =1 +cot^2x =csc^2x
sin A= -24/ 25 , tan B= -3/4 , cosB = -3/5
cos(A+B)=
cos AsinB +sinAcosB=(-7/25)(4/5)+(-24/25)(-3/5)=...,
sin2B =2sinBcosB=2 (4/5)(-3/5)= - 24/25
cos A/2 = sqrt [ (cosA+1)/2] = sqrt [ (-7/25+1)/2] = sqrt [9/25] = -3/5 ( notice A/2 in Q II)