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Trigonometry Practice Test 1

Question : 5 of 40

Marks: +1, -0

\frac{\tan x}{1+\sec x}-\frac{\tan x}{1-\sec x}

equal to?

Solution: 👈: Video Solution

Concept:

Simplify the difference of two fractions by taking the common denominator.

Explanation:

Let the given expression be

E

E=\frac{\tan x}{1+\sec x}-\frac{\tan x}{1-\sec x}

Take the LCM of the denominators:

E=\tan x\left(\frac{1-\sec x-(1+\sec x)}{(1+\sec x)(1-\sec x)}\right)

Simplify the numerator:

1-\sec x-1-\sec x=-2\sec x

Simplify the denominator using

1-\sec^2 x=-\tan^2 x

(1+\sec x)(1-\sec x)=1-\sec^2 x=-\tan^2 x

So,

E=\frac{\tan x(-2\sec x)}{-\tan^2 x}

Cancel

-\tan x

E=\frac{2\sec x}{\tan x}

Use

\sec x=\frac{1}{\cos x}

and

\tan x=\frac{\sin x}{\cos x}

E=\frac{2}{\sin x}=2\csc x

Answer:

2\csc x

Hence the correct option is B.

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