Explanation
since $$\sin x > 0$$ we can cancel it from both sides $$\cos ^{3}x > \cos x \sin^{2}x$$ $$ \cos x (\cos ^{2}x - \sin^{2}x) > 0 $$ $$ \cos x \cos 2x > 0 $$ which holds only if $$ 0 \leq x < \displaystyle \frac{\pi}{4}$$
$$|\tan x + \sec x| = |\tan x| - |\sec x|, x \epsilon [0,2\pi]$$if and only if x belongs to the interval
$$|\tan x + \sec x| = |\tan x| - |\sec x|$$
Then $$\tan x . \sec x \leq 0$$ and $$ |\tan x| \geq |\sec x|$$
$$Now, |\tan x| \geq |\sec x|$$ $$\Rightarrow \tan ^2x \geq \sec ^2x$$
$$\Rightarrow \tan ^2x \geq 1+ \tan ^2x \Rightarrow 0 \geq 1 $$ This is not possible.
$$\dfrac{\tan\alpha}{1-\cot\alpha}+\dfrac{\cot\alpha}{1-\tan\alpha}$$
$$=\dfrac{\tan\alpha}{1-\dfrac{1}{\tan\alpha}}+\dfrac{\cot\alpha}{1-\tan\alpha}$$
$$=\dfrac{\tan^2\alpha}{\tan\alpha-1}-\dfrac{\cot\alpha}{\tan\alpha-1}$$
$$=\dfrac{\tan^2-\dfrac{1}{\tan\alpha}}{\tan\alpha-1}$$
$$=\dfrac{\tan^3\alpha-1}{(\tan\alpha)(\tan\alpha-1)}$$
$$=\dfrac{(\tan\alpha-1)(\tan^2\alpha+\tan\alpha+1)}{(\tan\alpha)(\tan\alpha-1)}$$
$$=\dfrac{\tan^2\alpha+\tan\alpha+1}{\tan\alpha}$$
$$={\tan\alpha+1+\cot\alpha}$$
$$=\dfrac{\sin\alpha}{\cos\alpha}+\dfrac{\cos\alpha}{\sin\alpha}+1$$
$$=\dfrac{\sin^2\alpha+\cos^2\alpha}{\cos\alpha \sin\alpha}+1$$
$$=\dfrac{1}{\cos\alpha \sin\alpha}+1$$
$$=\sec\alpha\:\csc\alpha+1$$
Hence answer is C
We know that
$$\sec^2\beta-\tan^2\beta=1$$
Therefore
$$\tan\beta=\pm\sqrt{\sec^2\beta-1}$$
$$\tan\beta=\pm\sqrt{\alpha^2+(\dfrac{1}{4\alpha})\:^2+\dfrac{1}{2}}-1$$
$$\tan\beta=\pm\sqrt{\alpha^2+(\dfrac{1}{4\alpha})\:^2-\dfrac{1}{2}}$$
$$\tan\beta=\pm\sqrt{(\alpha-\dfrac{1}{4\alpha})\:^2}$$
$$\tan\beta=\pm(\alpha-\dfrac{1}{4\alpha})$$ ...(i)
Hence $$sec\alpha+\tan\alpha=2\alpha\:or\:\dfrac{1}{2\alpha}$$
The answer is B
$$ \cos x + \cos 2x + \cos 3x + \cos 4x + \cos 5x = 5$$ we know, $$ \cos x, \cos 2x, \cos 3x, \cos 4x, \cos 5x \leq 1 $$ considering both, $$ \cos x, \cos 2x, \cos 3x, \cos 4x, \cos 5x = 1$$ hence general solutions in the given interval are, $$ \cos x = 1 \Rightarrow x = 0, 2\pi$$ $$ \cos 2x = 1 \Rightarrow 2x = 0, 2\pi, 4\pi \Rightarrow x = 0, \pi, 2\pi $$ $$ \cos 3x = 1 \Rightarrow 3x = 0, 2\pi, 4\pi, 6\pi \Rightarrow x = 0, \dfrac{2\pi}{3}, 2\pi $$ $$ \cos 4x = 1 \Rightarrow 4x = 0, 2\pi, 4\pi, 6\pi, 8\pi \Rightarrow x = 0, \dfrac{\pi}{2},\pi, \dfrac{3\pi}{2}, 2\pi $$ $$ \cos 5x = 1 \Rightarrow 5x = 0, 2\pi, 4\pi, 6\pi, 8\pi, 10\pi \Rightarrow x = 0, \dfrac{2\pi}{5}, \dfrac{4\pi}{5}, \dfrac{6\pi}{5}, \dfrac{8\pi}{5}, 2\pi $$ thus common solutions are, $$ x = 0, 2\pi$$ Hence, option 'B' is correct.
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