If x is acute- then -1-sin x-1-sin xA- x-tan xB- cosec x- xC- x-cosec xD- tan x- x

The correct option is A sec x + tan xby rationalizing, √((1+sinx)*(1+sinx)/(1 sinx)*(1+sinx)) We get √((1+sinx)^2/(1 sinx)(1+sinx))=1+sinx/cosx=secx+tanx ...

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Standard X

Mathematics

Trigonometric Ratios

If x is acute...

Question

If x is acute, then $\sqrt{\frac{1 + s i n x}{1 - s i n x}}$

sec x + tan x

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cosec x + cot x

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sec x + cosec x

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tan x + cot x

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Solution

The correct option is A
sec x + tan x

by rationalizing, $\sqrt{\frac{(1 + s i n x) * (1 + s i n x)}{(1 - s i n x) * (1 + s i n x)}}$
We get $\sqrt{\frac{(1 + s i n x)^{2}}{(1 - s i n x) (1 + s i n x)}} = \frac{1 + s i n x}{c o s x} = s e c x + t a n x$

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If x is acute, then √1+sinx1−sinx

The correct option is A sec x + tan xby rationalizing, √(1+sinx)∗(1+sinx)(1−sinx)∗(1+sinx) We get √(1+sinx)2(1−sinx)(1+sinx)=1+sinxcosx=secx+tanx

If x is acute, then $\sqrt{\frac{1 + s i n x}{1 - s i n x}}$

The correct option is A
sec x + tan x

by rationalizing, $\sqrt{\frac{(1 + s i n x) * (1 + s i n x)}{(1 - s i n x) * (1 + s i n x)}}$
We get $\sqrt{\frac{(1 + s i n x)^{2}}{(1 - s i n x) (1 + s i n x)}} = \frac{1 + s i n x}{c o s x} = s e c x + t a n x$