Maths

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$\left(\frac{-1}{2}\right)$

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.

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)

54. Given, P (A) = P (B) where P is power set

Let x$\cancel{\in }$A.

Then, {x} $\subset$P (A)$\subset$ P (B)

i.e., x$\cancel{\in }$ B

A$\subset$ B

Similarly, B $\subset$A

A = B

53. Given, A $\subset$B.

Let x$\cancel{\in }$ C - B then x $\cancel{\in }$C but X$B$ .

However, A $\subset$B, elements of B should have elements of A

i.e., X$B$ 

So, x $\cancel{\in }$C A i.e., x$\cancel{\in }$ C but X$A$

C - B $\subset$C-  A

52. (i) Let A-  B≠$?$ , our assumption.

i.e., x$\in$ A But x≠ B where x is an element.

But as A B, the above condition of assumption is wrong if A $\subset$B then A -  B =$?$

(ii) Let x$\in$ A.

As A - B =$?$ we can say that x$\in$ B because if $x\cancel{\in }B,$ A - B ≠$?$

if A - B =$?$ then A $\subset$B.

(iii) We know that,

B $\subset$A$\cup$ B always true

Let x$\in$A$\cup$B i.e., x$\in$ A or x$\in$ B.

As A $\subset$B,

If x $\in$A then x$\in$ B, all elements of A are among the elements of B

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

(iv) We know that,

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

Let x $\in$A then x $\in$B

So, x $\in$(A$\cap$ B)

i.e., A$\subset$ (A$\cap$ B)

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

Hence, A$\subset$ B

A - B = $?$.

A$\cup$ B = B

A $\cap$B = A.

i.e., the 4 conditions are equivalent.

51. Let x$\in$b

As B$\subset$ A$\cup$ B we can write

Let x $\in$A $\cup$B.

as A $\cup$B = A $\cup$C.

x$\in$ A$\cup$ C.

i.e., x$\in$ A or x $\in$C

when x$\in$ A, and x$\in$ B,

x$\in$ A$\cap$ B

But A$\cap$ B = A $\cap$C

So, x$\in$ A $\cap$C

i.e., x$\in$ A and x$\in$ C

x $\in$C

when, x$\in$ C

as x $\in$B and x $\in$C

So, B $\subset$C

Similarly, C $\subset$B

So, B = C