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Illusion [34]
3 years ago
8

The expression (secx + tanx)2 is the same as _____.

Mathematics
1 answer:
trapecia [35]3 years ago
8 0

<u>Answer:</u>

The expression \bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}

<u>Solution:</u>

From question, given that \bold{(\sec x+\tan x)^{2}}

By using the trigonometric identity (a + b)^{2} = a^{2} + 2ab + b^{2} the above equation becomes,

(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x

We know that \sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}

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

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

On simplication we get

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

By using the trigonometric identity \cos ^{2} x=1-\sin ^{2} x ,the above equation becomes

=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}

By using the trigonometric identity (a+b)^{2}=a^{2}+2ab+b^{2}

we get 1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}

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

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

By using the trigonometric identity a^{2}-b^{2}=(a+b)(a-b)  we get 1-\sin ^{2} x=(1+\sin x)(1-\sin x)

=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}

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

Hence the expression \bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}

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