If \(x\) + y = 90°, then what is \(\sqrt{\text{cos}\,x\,cosec\,y\,-\,

1.	If \(x\) + y = 90°, then what is \(\sqrt{\text{cos}\,x\,cosec\,y\,-\,cos\,x\,sin\,y}\) equal to (a) cos \(x\) (b) sin \(x\) (c) \(\sqrt{cos\,x}\)(d) \(\sqrt{sin\,x}\)
Answer» (b) sin x \(x\) + y = 90° ⇒ y = 90° – \(x\) ∴ \(\sqrt{\text{cos}\,x\,cosec\,y\,-\,cos\,x\,sin\,y}\) = \(\sqrt{cox\,x\,cosec\,(90°-x)-cox\,x\,sin(90°-x)}\) = \(\sqrt{cos\,x\,sec\,x\,-cos\,x\,cos\,x}\) = \(\sqrt{1-cos^2\,x}=\sqrt{sin^2\,x}=sin\,x\)

If \(x\) + y = 90°, then what is \(\sqrt{\text{cos}\,x\,cosec\,y\,-\,cos\,x\,sin\,y}\) equal to (a) cos \(x\) (b) sin \(x\) (c) \(\sqrt{cos\,x}\)(d) \(\sqrt{sin\,x}\)

Answer»

(b) sin x

\(x\) + y = 90° ⇒ y = 90° – \(x\)

∴ \(\sqrt{\text{cos}\,x\,cosec\,y\,-\,cos\,x\,sin\,y}\)

= \(\sqrt{cox\,x\,cosec\,(90°-x)-cox\,x\,sin(90°-x)}\)

= \(\sqrt{cos\,x\,sec\,x\,-cos\,x\,cos\,x}\)

= \(\sqrt{1-cos^2\,x}=\sqrt{sin^2\,x}=sin\,x\)