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59. Let A, B and x be sets such that,

A$\cap$x = B$\cap$x =$?$ and A$\cup$x = B$\cup$x.

We know that,

A = A$\cap$ (A$\cup$x)

= A$\cap$ (B$\cup$x) [? A$\cup$x = $\cup$Bx]

= (A$\cap$B) (A$\cap$x) [by distributive law]

= (A$\cap$B) ∪? [? A∩x =? ]

=> A = A∩ B [? A ∪? = A]

And B = $\cap$ (B$\cup$x)

= B$\cap$ (A$\cup$x) [? B$\cup$x = A$\cup$x]

= (B$\cap$A)$\cup$ (B$\cap$x) [By distributive law]

= (B$\cap$A) ∪$?$ [? B$\cap$x =? ]

B = B$\cap$A [? A$\cup$ $?$= A]

So, A = B = A$\cap$B.

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.

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=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.

58. Let A = {a}, B = {a, b}, C = {a, c}

So, A$\cap$B = {a} $\cap$ {a, b} = {a}

A$\cap$C = {a} $\cap$ {a, c} = {a}

i.e., A$\cap$B = A$\cap$C = {a}

But B ≠C. as b$\cancel{\in }$B but $b\cancel{\ne }C$ vice-versa

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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.

57. (i) We know that,

A $\subset$A

(A $\cap$B)$\subset$ A

[A$\cup$ (A $\cap$B)] $\subset$ (A)

[A $\cup$ (A $\cap$B)] $\subset$A

and also

A$\subset$ [A $\cup$ (A $\cap$B)]

So, A$\cup$ (A$\cap$ B) = A.

(ii) A$\cap$ (A$\cup$ B) = (A$\cap$ A) $\cup$ (A$\cap$ B) [By distributive law]

= A$\cup$ (A$\cap$ B)

= A as (A $\cap$B) $\subset$A

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

= $\frac{\pi }{3}+2\times \frac{\pi }{6}$

= $\frac{\pi }{3}+\frac{\pi }{3}$

= $\frac{2\pi }{3}$

56. Here,

(A$\cap$B)$\cup$ (A - B) = (A$\cap$B)$\cup$ (A$\cap$ B’) as (A -  B) = A$\cap$ B’

= A $\cap$ (B$\cup$ B’) [ by converse of distributive law]

= A $\cap$U [ B $\cup$B’ = U, sample space set or universal set]

= A

And (A $\cup$ (B - A) = A $\cup$ (B$\cap$ A’) [as B -  A = B$\cap$ A’]

= (A$\cup$B) $\cap$ (A$\cup$A’)

= (A$\cup$B)$\cap$ U [ A$\cup$A’ = U, universal set]

= A$\cup$ B.

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

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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 $\left(-\frac{\pi }{6}\right)$

55. Let A = {a}, B = {b}.

So, P (A) = $\{?,\{a\left\}\right\}B(A)=\{?,(b\left\}\right\}$

So, P (A) $\cup$P {B} = $\left\{?,\left\{a\right\},\left\{b\right\}\right\}$ ______ (1)

Now, A$\cup$B = {a, b}.

P (A$\cup$B) = $\left\{?,\left\{a\right\},\left\{b\right\},\{a,b\}\right\}$ ____ (2)

So. From (1) and (2) we see that,

P (A)$\cup$ P (B) ≠P (A$\cup$B)

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