Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Hyperbolic Functions Solutions Exercise 9(a) will help students to clear their doubts quickly.

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

Question 1.

If sinh x = \(\frac{3}{4}\), find cosh (2x) and sinh (2x).

Solution:

Question 2.

If sinh x = 3, then show that x = log_{e}(3 + √10).

Solution:

Question 3.

Prove that

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

Solution:

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

Solution:

Question 4.

Prove that

(i) (cosh x – sinh x)^{n} = cosh (nx) – sinh (nx), for any n ∈ R.

Solution:

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

Solution:

Question 5.

Prove that \(\frac{\tanh x}{{sech} x-1}+\frac{\tanh x}{{sech} x+1}\) = -2 cosech x, for x ≠ 0

Solution:

Question 6.

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

Solution:

Question 7.

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

Solution:

L.H.S = 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]

= (1) cosh (2x)

= cosh (2x)

∴ cosh^{4}x – sinh^{4}x = 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.

Solution: