The product of the following series $(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + . . .)$ $\times$ $(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + . . .)$ is

Q. lf

α

lies in the third quadrant, then

\sqrt{\frac{1 - sin α}{1 + sin α}} + \sqrt{\frac{1 + sin α}{1 - sin α}} =

Q. Let

α

and

β

be the roots of equation

p x^{2} + q x + r = 0, p \neq 0

. If

p, q, r

are in A.P. and

\frac{1}{α} + \frac{1}{β} = 4

, then the value of

| α - β |

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The maximum value of cos(cosx) is

The correct option is D 1Maximum value of cos(cosx) will be obtained when value of cosx is minimum as cosθ is a decreasing FunctionSo, minimum value of cosx =0 (at x=90o,270o)Now, cos(cosx)=cos0=1Therefore, Answer is 1

The maximum value of $cos (cos x)$ is

The correct option is D $1$
Maximum value of $cos (cos x)$ will be obtained when value of $cos x$ is minimum as $cos θ$ is a decreasing Function

So, minimum value of $cos x = 0 (a t x = 90^{o}, 270^{o})$

Now, $cos (cos x) = cos 0 = 1$

Therefore, Answer is $1$