Answer
a) $\sin (\dfrac{\pi}{2}-u)=\cos u$
b) $\cos (\dfrac{\pi}{2}-u)=\sin u$
c) $\tan (\dfrac{\pi}{2}-u) =\cot u$
d) $\sec (\dfrac{\pi}{2}-u) =cosec u$
e) $cosec (\dfrac{\pi}{2}-u) =\sec u$
f) $\cot (\dfrac{\pi}{2}-u) =\tan u$
Work Step by Step
Here, we have $u+v+\dfrac{\pi}{2}=180^{\circ}$ or, $u+v+\dfrac{\pi}{2}=\pi$
This gives: $v=\dfrac{\pi}{2}-u$
Thus, we get
a) $\sin (\dfrac{\pi}{2}-u)=\sin (\dfrac{\pi}{2}) \cos u- \cos (\dfrac{\pi}{2}) \sin u=\cos u$
b) $\cos (\dfrac{\pi}{2}-u)=\cos (\dfrac{\pi}{2}) \cos u+ \sin (\dfrac{\pi}{2}) \sin u=\sin u$
c) $\tan (\dfrac{\pi}{2}-u) =\dfrac{\sin (\dfrac{\pi}{2}-u)}{\cos (\dfrac{\pi}{2}-u)}$
or, $\dfrac{\cos u}{\sin u}=\cot u$
Thus, $\tan (\dfrac{\pi}{2}-u) =\cot u$
d) $\sec (\dfrac{\pi}{2}-u) =\dfrac{1}{\cos (\dfrac{\pi}{2}-u)}$
or, $\dfrac{1}{\sin u}=cosec u$
Thus, $\sec (\dfrac{\pi}{2}-u) =cosec u$
e) $cosec (\dfrac{\pi}{2}-u) =\dfrac{1}{\sin (\dfrac{\pi}{2}-u)}$
or, $\dfrac{1}{\cos u}=\sec u$
Thus, $cosec (\dfrac{\pi}{2}-u) =\sec u$
f) $\cot (\dfrac{\pi}{2}-u) =\dfrac{\cos (\dfrac{\pi}{2}-u)}{\sin (\dfrac{\pi}{2}-u)}$
or, $\dfrac{\sin u}{\cos u}=\tan u$
Thus, $\cot (\dfrac{\pi}{2}-u) =\tan u$
