If - 2- x - then write the value of 1-cos 2 x 1-cos 2 x-

We have, 1 cos2x1+cos2x=2sin2x2cos2x=sinxcosx=sinx cosx #160; #160; #160; #160; #160; #8757; #960;2 #60;x #60; #960;= tanx ...

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Conditional Identities

If π 2< x <π,...

Question

If $\frac{π}{2} < x < π$ , then write the value of $\frac{\sqrt{1 - \cos 2 x}}{1 + \cos 2 x}$ .

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Solution

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f (x) = \{\begin{matrix} \frac{1 - \sin x}{π - 2 x}, & x \neq \frac{π}{2} \\ k, & x = \frac{π}{2} \end{matrix}

x = \frac{π}{2},

f (x) =

\frac{k cos x}{π - 2 x}

x \neq \frac{π}{2}

3

x = \frac{π}{2}

x = \frac{π}{2}

k

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - {sin}^{3} x}{3 {cos}^{2} x}, i f x < \frac{π}{2} a, i f x = \frac{π}{2} \frac{b (1 - sin x)}{(π - 2 x)^{2}}, i f x > \frac{π}{2} \end{matrix}

f (x)

x = \frac{π}{2}

a

b

f (x) = \frac{s i n (c o s x) - c o s x}{(π - 2 x)^{3}}, i f x \neq \frac{π}{2} = k, i f x = \frac{π}{2}

continuous at x = \frac{π}{2}, then k =

tan (π / 4 + x) + tan (π / 4 - x) = a

{tan}^{2} (π / 4 + x) + {tan}^{2} (π / 4 - x) = ?

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