If tan x = \(\frac{4}{3}\), then the value of \(\sqrt{\frac{(1-sin\,

1.	If tan x = \(\frac{4}{3}\), then the value of \(\sqrt{\frac{(1-sin\,x)(1+sin\,x)}{(1+cos\,x)(1-cos\,x)}}\) is(a) \(\frac{9}{16}\) (b) \(\frac34\)(c) \(\frac43\) (d) \(\frac{16}{9}\)
Answer» (b) \(\frac34\) \(\sqrt{\frac{(1-sin\,x)(1+sin\,x)}{(1+cos\,x)(1-cos\,x)}}\) = \(\sqrt{\frac{1-sin^2\,x}{1-cos^2\,x}}\) (∵ (a – b) (a+b) = a² – b²) = \(\sqrt{\frac{cos^2\,x}{sin^2\,x}}\) (∵ sin²x + cos²x = 1) = \(\frac{cos\,x}{sin\,x}\) = cot θ = \(\frac{1}{tan\,\theta} = \frac{1}{\frac{4}{3}}=\frac34.\)

If tan x = \(\frac{4}{3}\), then the value of \(\sqrt{\frac{(1-sin\,x)(1+sin\,x)}{(1+cos\,x)(1-cos\,x)}}\) is(a) \(\frac{9}{16}\) (b) \(\frac34\)(c) \(\frac43\) (d) \(\frac{16}{9}\)

Answer»

(b) \(\frac34\)

\(\sqrt{\frac{(1-sin\,x)(1+sin\,x)}{(1+cos\,x)(1-cos\,x)}}\) = \(\sqrt{\frac{1-sin^2\,x}{1-cos^2\,x}}\) (∵ (a – b) (a+b) = a² – b²)

= \(\sqrt{\frac{cos^2\,x}{sin^2\,x}}\) (∵ sin²x + cos²x = 1)

= \(\frac{cos\,x}{sin\,x}\) = cot θ = \(\frac{1}{tan\,\theta} = \frac{1}{\frac{4}{3}}=\frac34.\)