If u=log⌊tan(π /4+θ /2)⌋, then cosh u is

Correct Answer is : secθsec θsecθ

Correct Answer is : secθsec θsecθ

Test Index

AP DSC School Assistant Math Model Test 2

Section: Subject Specific

Question : 105 of 160

Marks: +1, -0

u=log⌊tan(\frac{\pi }{4}+\frac{\theta }{2})⌋

, then cosh u is

Solution:

Concept:

The hyperbolic cosine function is defined as

\cosh u = \frac{e^u + e^{-u}}{2}

. Given

u = \log\left[\tan\left(\frac{\pi}{4} + \frac{\theta}{2}\right)\right]

, we express

e^u

and

e^{-u}

using the tangent addition formula and simplify to find

\cosh u

in terms of

\theta

Explanation:

Let

t = \tan\frac{\theta}{2}

. Using the formula

\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}

with

A=\frac{\pi}{4}

B=\frac{\theta}{2}

, we have

\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right) = \frac{1+t}{1-t}

Thus

u = \log\frac{1+t}{1-t}

, so

e^u = \frac{1+t}{1-t}

and

e^{-u} = \frac{1-t}{1+t}

Now compute

e^u + e^{-u}

e^u+e^{-u} = \frac{1+t}{1-t} + \frac{1-t}{1+t} = \frac{(1+t)^2+(1-t)^2}{(1-t)(1+t)} = \frac{(1+2t+t^2)+(1-2t+t^2)}{1-t^2} = \frac{2+2t^2}{1-t^2} = \frac{2(1+t^2)}{1-t^2}

Therefore,

\cosh u = \frac{e^u+e^{-u}}{2} = \frac{1+t^2}{1-t^2}

Substitute

t = \tan\frac{\theta}{2}

and use the identities

1+\tan^2\frac{\theta}{2} = \sec^2\frac{\theta}{2}

and

1-\tan^2\frac{\theta}{2} = \frac{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}{\cos^2\frac{\theta}{2}} = \frac{\cos\theta}{\cos^2\frac{\theta}{2}}

. Then

\cosh u = \frac{\sec^2\frac{\theta}{2}}{\frac{\cos\theta}{\cos^2\frac{\theta}{2}}} = \sec^2\frac{\theta}{2} \cdot \frac{\cos^2\frac{\theta}{2}}{\cos\theta} = \frac{1}{\cos\theta} = \sec\theta

Answer:

\cosh u = \sec\theta

(Option A)

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