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Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$

My attempt,

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

$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$

$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$

I'm stuck at here. The given answer is $\sec x -\sin x$

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3 Answers 3

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Hint

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

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

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    $\begingroup$ this gives $$\sec(x)-\sin(x)$$ $\endgroup$ Jun 19, 2017 at 15:10
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    $\begingroup$ @Dr.SonnhardGraubner: that seems to be the answer. $\endgroup$
    – Arnaldo
    Jun 19, 2017 at 15:12
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From where you were stuck,
$$\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)} = \frac{\cos^2x(1 - \sin^2x) + \sin^2x}{\cos x(\sin x \cos x+1)} = \frac{(\cos^2x + \sin^2x) - \cos^2x\sin^2x}{\cos x(\sin x \cos x+1)} = \frac{1 - \cos^2x\sin^2x}{\cos x(\sin x \cos x+1)} = \frac{(1+\sin x \cos x)(1-\sin x \cos x)}{\cos x(\sin x \cos x+1)} = \frac{1-\sin x\cos x}{\cos x}$$ $$=\sec x -\sin x$$

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$$\scriptsize\frac {\tan^2x+\color{blue}{\cos^2x}}{\sin x+\sec x}=\frac {\overbrace{\tan^2x+\color{blue}1}^{\sec^2 x}\color{blue}{-\sin^2x}}{\sin x+\sec x}=\frac {\sec^2x-\sin^2x}{\sec x+\sin x}=\frac{(\sec x -\sin x)(\sec x+\sin x)}{\sec x+\sin x}=\sec x-\sin x$$

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