# Trigonometric proof involving sec and cosec

How should I proceed to prove the following: $$\sec^2 A +\csc^2 A = \sec^2 A \cdot \csc^2 A$$

This is what I've attempted:

$$LHS = 1 + \tan^2 A + 1 + \cot^2= 2+ 1\cot^2 + \cot^2$$

However, this gets me nowhere. How would it be done?

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If you replace the $-$ by a $+$ on the lhs, it will be easier: common denominator and $\cos^2A+\sin^2A=1$. –  1015 Apr 19 '13 at 13:08
@julien Oh yes thanks for pointing that out. I've changed it. –  Ghost Apr 19 '13 at 13:18

(You have a sign error.) Personally I'd expand in $\sin,\cos$.

Hint: You find

$$\sec^2 + \csc^2 = \frac{1}{\cos^2} + \frac{1}{\sin^2} = \frac{\cdots}{\sin^2\cos^2}$$

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$$\sec^2A\csc^2A-\sec^2A=\sec^2A(\csc^2A-1)=\sec^2A\cot^2A=\csc^2A$$

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Just to mention it, here's a geometric approach that uses a depiction of the trig values as lengths of segments of a triangle (which I call the "Complete Triangle"):

Compute twice the area of the big triangle using "base-times-height" in two ways:

$$\sec\theta \cdot \csc\theta = \left(\tan\theta+\cot\theta\right)\cdot 1$$

Now, square both sides, and observe that $\tan\theta+\cot\theta$ is the length of the hypotenuse of the big triangle:

$$\sec^2\theta \cdot \csc^2\theta = \left( \tan\theta + \cot\theta \right)^2 = \sec^2\theta + \csc^2\theta$$

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it shud not be complicated –  Mr.ØØ7 Apr 19 '13 at 13:53