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Question

Find the minimum value of $$\displaystyle \left| sin x + cos x + \frac{cos x - sin x}{cos^4 x - sin^4 x} \right|$$


Solution

Let $$y=\left |\sin x+\cos x+\dfrac {\cos x-\sin x}{\cos^4x - \sin^4 x}\right|....(1)$$
we have to find minimum value of $$y$$
$$\Rightarrow \ y=\left |\sin x+\cos x+\dfrac {\cos x-\sin x}{(\cos^2x +\sin^2x) (\cos^2x-\sin^2x)}\right|$$
$$\Rightarrow \ y=\left |\sin x+\cos x+\dfrac {\cos x-\sin x}{(\cos x-\sin x) (\cos x+\sin x)} \right|$$
$$\Rightarrow \ y=\left | \dfrac {(\sin x+\cos x)^2+1}{\cos x+\sin x}\right|$$...... on expanding
$$y=\left |\dfrac {2+2\sin x \cos x}{\cos x+\sin x} \right|......(2)$$
$$\Rightarrow $$ In order to minimize $$y=$$ we have to maximize denominator Multyply & divide by $$\sqrt 2$$
$$\sqrt 2 \left (\dfrac {\cos x}{\sqrt 2}+\dfrac {\sin x}{\sqrt 2}\right) \Rightarrow \sqrt 2 \left (\sin \left (x+\dfrac {\pi}{4}\right) \right)$$
now equation $$(2)$$ becomes
$$y=\left |\dfrac {\sqrt 2 (1+\sin x\cos x)}{\sin \left (x+\dfrac {\pi}{4}\right)}\right|.....(3)$$
we know $$\sin \left (x+\dfrac {\pi}{4}\right) \le 1$$
$$x+\dfrac {\pi}{4}\le \dfrac {\pi}{2} \Rightarrow \left [x=\dfrac {\pi}{4}\right]$$
$$\Rightarrow \ $$ Hence at $$x=\dfrac {pi}{4}$$ value of $$y$$ in minimum
$$\Rightarrow \ $$ now from equation $$(3)$$
$$y=\left |\dfrac {\sqrt 2 \left (1+\dfrac {1}{\sqrt 2}\times \dfrac {1}{\sqrt 2} \right)}{\sin \left (\dfrac {\pi}{2}\right)}\right|$$
$$y=\sqrt  2\left (\dfrac {3}{2}\right)=\dfrac {3}{\sqrt 2}$$
$$\Rightarrow \ \left (y=\dfrac {3}{\sqrt 2}\right)$$

Mathematics

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