Class 12th

We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3 (cos^-1x) = 4x³-3x

3cos^-1x = cos^-1 (4x³- 3x)

Hence Proved

We know that.

sin 3θ =3 sin θ 4sin³θ (identity).

(E) Let x = sinθ. Then, sin ⁻¹x=θ . We have,

Sin3 (sin ⁻¹x) = 3x−4x³

3sin ⁻¹x =sin^-1 (3x−4x³)

Hence proved.

=tan⁻¹ $\left(\frac{tan\pi }{3}\right)$ − $\pi$ + sec⁻¹$\left(sec\frac{\pi }{3}\right)$

= $\frac{\pi }{3}-\pi +\frac{\pi }{3}$

= $\frac{\pi -3\pi +\pi }{3}$

= $\frac{-\pi }{3}.$

Option B is correct.

Given, Sin⁻¹x=y.

(E) We know that the principal value branch of Sin⁻¹ is

$[\frac{-\pi }{2},\frac{\pi }{2}]$ Hence,  $\frac{-\pi }{2}$ ≤ y ≤ $\frac{\pi }{2}$

Option B is correct.

cos⁻¹$\frac{1}{2}$ + 2Sin⁻¹ $\frac{1}{2}$ = cos⁻¹ $\left(cos\frac{\pi }{3}\right)$ + 2*Sin⁻¹$\left(sin\frac{\pi }{6}\right)$

= $\frac{\pi }{3}+2*\frac{\pi }{6}$

= $\frac{\pi }{3}+\frac{\pi }{3}$

= $\frac{2\pi }{3}$

tan⁻¹ (1) + cos⁻¹ $\left(\frac{1}{2}\right)$ +  sin ⁻¹$\left(\frac{-1}{2}\right)$

Let cos ^-1$\left(\frac{-1}{2}\right)$ =y Then cos y = $\frac{-1}{2}$ = − cos
$\frac{\mathrm{}\pi }{3}$$\frac{}{}$ cos $\left(\pi -\frac{x}{3}\right)$

= cos $\frac{3\pi -\pi }{3}$

= cos $\frac{2\pi }{3}$

$$  (E) We know that the range of principal value

branch of cos⁻¹ is [0, $\frac{\pi }{}$] and cos $\frac{2x}{3}$ = $\frac{-1}{2}$

Principal value of cos−1 $\left(\frac{-1}{2}\right)$ is $\frac{2x}{3}$

Let sin⁻¹ $$ $\left(\frac{-1}{2}\right)$ =y. Then, sin y=- $\frac{-1}{2}$

We know that the range of principal value branch of sin⁻¹ is $-\frac{\pi }{2},\;$

and sin⁻¹ $\left(\frac{-1}{2}\right)$ =−sin⁻¹$\frac{-1}{2}$ (sin (-x) = -sin x)

= $\left(-\frac{\pi }{6}\right)$ = sin y (as sin
$\frac{\pi }{6}$
$\frac{}{}$ = $\frac{1}{2}$ )

Principal value of sin⁻¹ $\left(-\frac{1}{2}\right)$ is $\left(-\frac{\pi }{6}\right)$

4.39 Given, k₂ = 4k₁, T₁ = 293K and T₂ = 313K

We know that from the Arrhenius equation, we obtain

On solving, we get,

E_a = 58263.33 J mol^-1 or 58.26 kJ mol^-1