# Bihar Board 12th Maths Important Questions Long Answer Type Part 2

BSEB Bihar Board 12th Maths Important Questions Long Answer Type Part 2 are the best resource for students which helps in revision.

## Bihar Board 12th Maths Important Questions Long Answer Type Part 2

Determinants

Question 1.
Using the property of determinants and without expending, prove that
(i) $$\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|$$ = 0
(ii) $$\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$$ = 0
(iii) $$\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & a(c+a) \\ 1 & a b & c(a+b) \end{array}\right|$$ = 0
Solution:

Question 2.
By using determinants property $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|$$ = (a- b)(b – c)(c – a)(a + b + c)
Solution:

= (a-b) (b-c)-(b2 +bc+c2 -a2-ab-b2)
= (a-b)-(b-c)-{(c2-a2) + b-(c-a)}
= (a-b)(b-c)-{(c-a)(c + a) + b-(c-a)}
= (a-b)(b-c)-(c-a)(c + a + b)
= (a-b)(b-c)-(c-a)-(a + b+c) R.H.S.

Question 3.
Prove that $$\left|\begin{array}{ccc} a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 h+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c \end{array}\right|=a^{2}$$
Solution:
L.H.S = $$\left|\begin{array}{lll} a & a+b & a+b+c \\ 2 a & 3 a+2 b & 4 a+3 b+2 c \\ 3 a & 6 a+3 b & 10 a+6 b+3 c \end{array}\right|$$
R2 → R2 – 2R3, R3 → R3 – 3R1
= $$\left|\begin{array}{rrr} a & a+b & a+b+c \\ 0 & a & 2 a+b \\ 0 & 3 a & 7 a+3 b \end{array}\right|$$
From 1st Column
= a$$\left|\begin{array}{cc} a & 2 a+b \\ 3 a & 7 a+3 b \end{array}\right|$$
= a(7a2 + 3ab + 6a2 – 3ab)
= a.a2 = a3 = RHS

Question 4.
Without expanding prove that
$$\left|\begin{array}{ccc} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{array}\right|$$ = 0
Solution:
L.H.S = $$\left|\begin{array}{ccc} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{array}\right|$$
R1→ R1 + R2
= $$\left|\begin{array}{ccc} x+y+z & y+z+x & z+x+y \\ z & x & y \\ 1 & 1 & 1 \end{array}\right|$$
= (x + y + z)$$\left|\begin{array}{lll} 1 & 1 & 1 \\ z & x & x \\ 1 & 1 & 1 \end{array}\right|$$
= (x + y + z) = 0 = 0 R.H.S.

Question 5.
Evaluate : $$\left|\begin{array}{lll} 1 & a & b c \\ 1 & b & c a \\ 1 & c & a b \end{array}\right|$$
Solution:

= (a -b)(b – c)(-a + c)
= (a – b) (b – c) (c – a) = R.H.S

Question 6.
Without expanding shw that,
$$\left|\begin{array}{lll} a+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|=2\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & 2 \end{array}\right|$$
Solution:

Question 7.
Prove that the determinants $$\left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right|$$ is independent of θ
Solution:
Let Δ = $$\left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right|$$
From 1st row→
= $$x \cdot\left|\begin{array}{cc} -x & 1 \\ 1 & x \end{array}\right|-\sin \theta\left|\begin{array}{cc} -\sin \theta & 1 \\ \cos \theta & x \end{array}\right|+\cos \theta\left|\begin{array}{rr} -\sin \theta & -x \\ \cos \theta & 1 \end{array}\right|$$
= x(-x2 – 1) – sinθ.(-xsinθ – cos θ) + cos θ(-sin θ + x cos θ)
= -x3 – x + xsin2θ + sin θ.cos θ – sin θ.cos θ + xcos2θ
= -x3.x + x.(sin2θ + cos2θ)
= -x3 – x + x.1
= -x3 – x + x
= -x3 = indepentdent

Question 8.
Evaluate :
$$\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$$ = (1 – x2)2
L.H.S = Δ = $$\left|\begin{array}{ccc} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$$
Operating C1 → C1 + C2 + C3

= (1 – x)2(1 + x + x2)(1) + x(1 + x)
= (1 – x)2 (1 + x + x2)[1 + x + x2]
= (1 – x)2(1 + x + x2 )2
= [(1 – x)(1 + x + x2 )]2 = (1 – x3)2

Question 9.
Evaluate :
$$\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}$$

Expanding we get
Δ = (1 + a +b2)2 (1 – a2 – b2) + 2a2 + 2b2
= (1 + a2 + b2 )2 (1 + a2 + b2) = (1 + a2 + b2)3

Question 10.
Evaluate :
$$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$$ = 1 + a2 + b2 + c2
Solution:

This may be expressed as the sum of 8 determinants.

= abc  + (b2c2 – b2c2) + (a2c2 – a2c2) + (a2b2 – a2b2) + a2 +b2 + c2
= a2 + b2 + c2

Inverse Circular Function

Question 1.
Show the following :
(i) sin-1(2x$$\sqrt{1-x^{2}}$$) = 2sin-1x
(ii) sin-1$$\frac { 8 }{ 17 }$$ + -1$$\frac { 3 }{ 5 }$$ = -1$$\frac { 77 }{ 85 }$$
(iii) sin [cot-1(cos(tan-1x)}] = $$\frac{\sqrt{1+x^{2}}}{2+x^{2}}$$
(iv) 2tan-1 $$\frac { 1 }{ 2 }$$ + tan-1 $$\frac { 1 }{ 7 }$$ = tan-1$$\frac { 31 }{ 17 }$$
(v) $$\tan ^{-1} \sqrt{x} \frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$$ , x ∈ [0,1].
Solution:
(i) Let sin-1 = θ
x = sin θ
L.H.S. = sin-1 ($$2 x \cdot \sqrt{1-x^{2}}$$)
= sin-1(2sinθ $$\sqrt{1-\sin ^{2} \theta}$$)
= sin-1(2sinθ.cosθ) = sin-1(sin2 θ)
= 2θ = 2sin-1 x = R.H.S.

(ii) L.H.S. = sin-1 $$\frac{8}{17}$$ + sin-1 $$\frac{3}{5}$$

(iii) L.H.S. = sin[cot-1 {cos(tan-1 x)}]
Let tan-1 x = y ⇒ A = tan y
= sin[cot -1(cosy)]

(iv)

(v) L.H.S = $$\tan ^{-1} \sqrt{x}=\frac{1}{2} \cdot 2 \tan ^{-1} \sqrt{x}$$
= $$\frac{1}{2} \cdot \cos ^{-1}\left(\frac{1-(\sqrt{x})^{2}}{1+(\sqrt{x})^{2}}\right)=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$$ = R.H.S

Question 2.
Find the values of each of the following :
(i) $$\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]$$
(ii) $$\tan \frac{1}{2} \cdot\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]$$, |x| < 1, y > 0, x,y < 1
Solution:

(ii) $$\tan \frac{1}{2} \cdot\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]$$
= tan $$\tan \frac{1}{2}$$ [2tan-1 x + 2tan-1 y]
= tan[tan-1x + tan-1 y]
= tan [tan-1( $$\frac{x+y}{1-x y}$$ )
= $$\frac{x+y}{1-x y}$$

Question 3.
Show that $$\cot ^{-1}\left(\frac{\sqrt{1+\sin } \bar{x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\mu}{4}\right)$$
Show that $$\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}$$