Question #8331a

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Question #8331a

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Answer 1 · 2016-03-20T12:20:36

As below

Explanation:

To Prove ${(tan x + cot x)}^{4} = {csc}^{4} x \cdot {sec}^{4} x$ $(tanx+cotx)^4=csc^4 x cdot sec^4 x$
LHS $= {(tan x + cot x)}^{4}$ $=(tanx+cotx)^4$
write $tan x and cot x$ $tanx and cot x$ in terms of $sin and cos$ $sin and cos$
LHS $= {(\frac{sin x}{cos x} + \frac{cos x}{sin x})}^{4}$ $=(sinx/cosx+cosx/sinx)^4$ , simplify
$\Rightarrow {(\frac{sin x \times sin x + cos x \times cos x}{cos x \cdot sin x})}^{4}$ $=> ((sinx xxsinx+cosx xxcosx)/(cosx cdot sinx))^4$
$\Rightarrow {(\frac{{sin}^{2} x + {cos}^{2} x}{cos x \cdot sin x})}^{4}$ $=> ((sin^2x +cos^2x)/(cosx cdot sinx))^4$ , Use Identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x +cos^2x=1$
$\Rightarrow {(\frac{1}{cos x \cdot sin x})}^{4}$ $=> (1/(cosx cdot sinx))^4$ by definition of $sec and csc$ $sec and csc$
$\Rightarrow {(sec x \cdot csc x)}^{4}$ $=> (sec x cdotcscx)^4$
$\Rightarrow {sec}^{4} x \cdot {csc}^{4} x =$ $=> sec^4 x cdotcsc^4x =$ RHS

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Question #8331a

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