Question

If cot theta = 7/8 , evaluate (I) cos square theta + sin square theta (Ii) cos square theta - sin square theta

Answer 1

cotθ=7/8=B/P. By pythagoras theorem, H=P2+B2=82+72=113. Now,cosθ=BH=7113sinθ=PH=81131)cos2θ +sin2θ=71132+81132=49+64113=113113=12)cos2θ- sin2θ=71132-81132=49-64113=-15113

Answer 2

cot theta = B/P =  7/8 
By pythagoras theorem,
H=√P2 + B2
H= √8² + 7²
H= √64+49
H= √113
 1) cos²theta + sin²theta
cos theta = B/H = 7/√113
sin theta = P/H = 8/√113
by putting values we get,
= (7/√113)² +(8/√113)²
= (49 + 64)/113 = 113/113 = 1 

2) cos²theta - sin²theta
= (7/√113)² - (8/√113)²
= (49-64)/113 = -15/113 = -0.13