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Trigonometry Practice Test 1

Question : 9 of 40

Marks: +1, -0

p=\sqrt{\frac{1-\sin x}{1+\sin x}}, q=\frac{1-\sin x}{\cos x}

r=\frac{\cos x}{1+si x}

Which one of the following is correct?

Solution: 👈: Video Solution

Concept:

Use

1-\sin^2 x=\cos^2 x

and rationalization to compare the three expressions.

Explanation:

Given

p=\sqrt{\frac{1-\sin x}{1+\sin x}}

Multiply numerator and denominator inside the root by

(1-\sin x)

p=\sqrt{\frac{(1-\sin x)^2}{1-\sin^2 x}}

Since

1-\sin^2 x=\cos^2 x

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

For the usual domain

\cos x>0

p=\frac{1-\sin x}{\cos x}=q

Now rationalize

q

q=\frac{1-\sin x}{\cos x}\times\frac{1+\sin x}{1+\sin x}

q=\frac{1-\sin^2 x}{\cos x(1+\sin x)}

q=\frac{\cos^2 x}{\cos x(1+\sin x)}

q=\frac{\cos x}{1+\sin x}=r

Hence

p=q

and

q=r

, so

p=q=r

Answer:

Option D:

p=q=r

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