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Question

Question: How do you simplify \[\sin x\sec x\]?...

How do you simplify sinxsecx\sin x\sec x?

Explanation

Solution

Here, we will assume the angle xx to be the interior angle of a right angled triangle. We will express the trigonometric ratios in terms of sides of the right angled triangle. Then we will simplify the given expression by multiplying the terms to get the required answer.

Complete step-by-step answer:
Let angle xx be the interior angle of a right angled triangle with P as perpendicular, B as base, and H as hypotenuse.
The sine of an angle of a right angled triangle is the ratio of its perpendicular and hypotenuse.
Therefore, we get
sinx=PH\sin x = \dfrac{P}{H}

The secant of an angle of a right angled triangle is the ratio of its hypotenuse and base.
Therefore, we get
secx=HB\sec x = \dfrac{H}{B}
Substituting sinx=PH\sin x = \dfrac{P}{H} and secx=HB\sec x = \dfrac{H}{B} in the given expression, we get
sinxsecx=PH×HB\sin x\sec x = \dfrac{P}{H} \times \dfrac{H}{B}
Simplifying the expression, we get
sinxsecx=PB\Rightarrow \sin x\sec x = \dfrac{P}{B}
The tangent of an angle of a right angled triangle is the ratio of its perpendicular and base.
Therefore, we get
tanx=PB\tan x = \dfrac{P}{B}
Substituting PB=tanx\dfrac{P}{B} = \tan x in the equation sinxsecx=PB\sin x\sec x = \dfrac{P}{B}, we get
sinxsecx=tanx\Rightarrow \sin x\sec x = \tan x
Therefore, we have simplified the expression sinxsecx\sin x\sec x as tanx\tan x.

Note: We can also convert the trigonometric ratio of secant to cosine, and then the quotient of sine and cosine to tangent to simplify the given expression.
The secant of an angle xx can be written as the reciprocal of cosine of the angle xx. This can be written as secx=1cosx\sec x = \dfrac{1}{{\cos x}}.
Substituting secx=1cosx\sec x = \dfrac{1}{{\cos x}} in the given expression, we get
sinxsecx=sinx×1cosx\Rightarrow \sin x\sec x = \sin x \times \dfrac{1}{{\cos x}}
Rewriting the expression, we get
sinxsecx=sinxcosx\Rightarrow \sin x\sec x = \dfrac{{\sin x}}{{\cos x}}
The secant of an angle xx can be written as the quotient of the sine and cosine of the angle xx. This can be written as tanx=sinxcosx\tan x = \dfrac{{\sin x}}{{\cos x}}.
Substituting sinxcosx=tanx\dfrac{{\sin x}}{{\cos x}} = \tan x in the equation sinxsecx=sinxcosx\sin x\sec x = \dfrac{{\sin x}}{{\cos x}}, we get
sinxsecx=tanx\Rightarrow \sin x\sec x = \tan x
Therefore, we have simplified the expression sinxsecx\sin x\sec x as tanx\tan x.