If p sin x = q and x is acute then \(\sqrt{p^2-q^2}\) tan x is equal t

1.	If p sin x = q and x is acute then \(\sqrt{p^2-q^2}\) tan x is equal to(a) p (b) q (c) pq (d) p + q
Answer» (b) q \(\sqrt{p^2-q^2}\) tan \(x\) = \(\sqrt{p^2-q^2\,sin^2\,x.}\,tan\,x\) = \(\sqrt{p^2(1-sin^2\,x).}\) tan \(x\) = \(p\sqrt{1-sin^2\,x}.\,tan\,x\) (∵cos²θ + sin²θ = 1) = \(p\sqrt{cos^2\,x.}\) tan \(x\) = p cos \(x\). tan \(x\) = p sin \(x\) = q

If p sin x = q and x is acute then \(\sqrt{p^2-q^2}\) tan x is equal to(a) p (b) q (c) pq (d) p + q

Answer»

(b) q

\(\sqrt{p^2-q^2}\) tan \(x\) = \(\sqrt{p^2-q^2\,sin^2\,x.}\,tan\,x\)

= \(\sqrt{p^2(1-sin^2\,x).}\) tan \(x\) = \(p\sqrt{1-sin^2\,x}.\,tan\,x\) (∵cos²θ + sin²θ = 1)

= \(p\sqrt{cos^2\,x.}\) tan \(x\) = p cos \(x\). tan \(x\) = p sin \(x\) = q