Students must practice these TS Intermediate Maths 1A Solutions Chapter 9 Hyperbolic Functions Ex 9(a) to find a better approach to solving the problems.

## TS Inter 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a)

I.

Question 1.

If sinh x = \(\frac{3}{4}\) find cosh (2x) and sinh (2x). (May 2014, Mar.’14, ’12)

Answer:

Given sinh x = \(\frac{3}{4}\)

and we have cosh^{2} x – sinh^{2} x = 1

Question 2.

If sinhx = 3, then show that

x = log_{e}(3 + √10) (Board New Model Paper).

Answer:

Given sin hx = 3

⇒ x = sinh^{-1} 3 = log_{e} (3 + \(\sqrt{3^2+1}\))

= log_{e} (3 + √10)

(∵ sinh x = log (x + \(\sqrt{x^2+1}\)) ∀ x ∈ R)

Question 3.

Prove that

(i) tanh (x – y) = \(\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)

Answer:

(ii) coth (x – y) = \(\frac{{coth} x \cdot {coth} y-1}{{coth} y-{coth} x}\)

Answer:

Question 4.

Prove that

(i) (cosh x – sinh x)^{n}

= cosh (nx) – sinh (nx) for any n ∈R.

Answer:

(ii) (cosh x + sinh x)^{n} = cosh (nx) + sinh (nx) for any n ∈ R.

Answer:

Question 5.

Prove that

\(\frac{\tanh x}{{sech} x-1}+\frac{\tanh x}{{sech} x+1}\) = -2cosechx for x ≠ 0.

Answer:

Question 6.

Prove that \(\frac{\cosh x}{1-\tanh x}+\frac{\sinh x}{1-{coth} x}\) = sinhx + coshx for x ≠ 0.

Answer:

Question 7.

For any x ∈ R, prove that cosh^{4} x – sinh^{4} x = cosh 2x.

Answer:

cosh^{4} x – sinh^{4} x

= (cosh^{2} x)^{2} – (sinh^{2} x)^{2}

= (cosh^{2} x + sinh^{2} x) (cosh^{2} x – sinh^{2} x)

= cosh 2x (1) = cosh 2x

Question 8.

If u = log_{e} \(\left\{\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right\}\) and if cos θ > 0 then prove that cosh u = sec θ

Answer: