Maharashtra Board 12th Maths Solutions Chapter 3 Trigonometric Functions Miscellaneous Exercise 3

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Miscellaneous Exercise 3 Questions and Answers.

Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Miscellaneous Exercise 3

I) Select the correct option from the given alternatives.
Question 1.
The principal of solutions equation sinθ = $\frac{-1}{2}$ are ________.

Solution:
(b) $\frac{7 \pi}{6}, \frac{11 \pi}{6}$

Question 2.
The principal solution of equation cot θ = $\sqrt {3}$ ___________.

Solution:
(a) $\frac{\pi}{6}, \frac{7 \pi}{6}$

Question 3.
The general solution of sec x = $\sqrt {2}$ is __________.
(a) 2nπ ± $\frac{\pi}{4}$, n ∈ Z
(b) 2nπ ± $\frac{\pi}{2}$, n ∈ Z
(c) nπ ± $\frac{\pi}{2}$, n ∈ Z
(d) 2nπ ± $\frac{\pi}{3}$, n ∈ Z
Solution:
(a) 2nπ ± $\frac{\pi}{4}$, n ∈ Z

Question 4.
If cos pθ = cosqθ, p ≠ q rhen ________.
(a) θ = $\frac{2 n \pi}{p \pm q}$
(b) θ = 2nπ
(c) θ = 2nπ ± p
(d) nπ ± q
Solution:
(a) θ = $\frac{2 n \pi}{p \pm q}$

Question 5.
If polar co-ordinates of a point are $\left(2, \frac{\pi}{4}\right)$ then its cartesian co-ordinates are ______.
(a) (2, $\sqrt {2}$ )
(b) ($\sqrt {2}$, 2)
(c) (2, 2)
(d) ($\sqrt {2}$ , $\sqrt {2}$)
Solution:
(d) ($\sqrt {2}$ , $\sqrt {2}$)

Question 6.
If $\sqrt {3}$ cosx – sin x = 1, then general value of x is _________.
(a) 2nπ ± $\frac{\pi}{3}$
(b) 2nπ ± $\frac{\pi}{6}$
(c) 2nπ ± $\frac{\pi}{3}-\frac{\pi}{6}$
(d) nπ + (-1)ⁿ$\frac{\pi}{3}$
Solution:
(c) 2nπ ± $\frac{\pi}{3}-\frac{\pi}{6}$

Question 7.
In ∆ABC if ∠A = 45°, ∠B = 60° then the ratio of its sides are _________.
(a) 2 : $\frac{\pi}{2}$ : $\frac{\pi}{3}$ + 1
(b) $\frac{\pi}{2}$ : 2 : $\frac{\pi}{3}$ + 1
(c) 2 $\frac{\pi}{2}$ : $\frac{\pi}{2}$ : $\frac{\pi}{3}$
(d) 2 : 2 $\frac{\pi}{2}$ : $\frac{\pi}{3}$ + 1
Solution:
(a) 2 : $\frac{\pi}{2}$ : $\frac{\pi}{3}$ + 1

Question 8.
In ∆ABC, if c² + a² – b² = ac, then ∠B = __________.
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{2}$
(d) $\frac{\pi}{6}$
Solution:
(b) $\frac{\pi}{3}$

Question 9.
In ABC, ac cos B – bc cos A = ____________.
(a) a² – b²
(b) b² – c²
(c) c² – a²
(d) a² – b² – c²
Solution:
(a) a² – b²

Question 10.
If in a triangle, the are in A.P. and b : c = $\sqrt {3}$ : $\sqrt {2}$ then A is equal to __________.
(a) 30°
(b) 60°
(c) 75°
(d) 45°
Solution:
(c) 75°

Question 11.
cos^-1$\left(\cos \frac{7 \pi}{6}\right)$ = ________.

Question 12.
The value of cot (tan^-1 2x + cot^-1 2x) is __________.
(a) 0
(b) 2x
(c) π + 2x
(d) π – 2x
Solution:
(a) 0

Question 13.
The principal value of sin^-1$\left(-\frac{\sqrt{3}}{2}\right)$ is ____________.

Solution:
(d) $-\frac{\pi}{3}$

Question 14.
If sin^-1$\frac{4}{5}$ + cos^-1$\frac{,12}{13}$ = sin^-1 ∝, then ∝ = _____________.
(a) $\frac{63}{65}$
(b) $\frac{62}{65}$
(c) $\frac{61}{65}$
(d) $\frac{60}{65}$
Solution:
(a) $\frac{63}{65}$

Question 15.
If tan^-1(2x) + tan^-1(3x) = $\frac{\pi}{4}$, then x = ________.
(a) -1
(b) $\frac{1}{6}$
(c) $\frac{2}{6}$
(d) $\frac{3}{2}$
Solution:
(b) $\frac{1}{6}$

Question 16.
2 tan^-1$\frac{1}{3}$ + tan^-1$\frac{1}{7}$ = ______.
(a) tan^-1$\frac{4}{5}$
(b) $\frac{\pi}{2}$
(c) 1
(d) $\frac{\pi}{4}$
Solution:
(d) $\frac{\pi}{4}$

Question 17.
tan (2 tan^-1$\left(\frac{1}{5}\right)-\frac{\pi}{4}$) = ______.
(a) $\frac{17}{7}$
(b) $-\frac{17}{7}$
(c) $\frac{7}{17}$
(d) $-\frac{7}{17}$
Solution:
(d) $-\frac{7}{17}$

Question 18.
The principal value branch of sec^-1 x is __________.

Solution:
(b) [0, π] – {$\frac{\pi}{2}$}

Question 19.
cos[tan^-1$\frac{1}{3}$ + tan^-1$\frac{1}{2}$] = ________.
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) $\frac{\pi}{4}$
Solution:
(a) $\frac{1}{\sqrt{2}}$

Question 20.
If tan θ + tan 2θ + tan 3θ = tan θ∙tan 2θ∙tan 3θ, then the general value of the θ is _______.
(a) nπ
(b) $\frac{n \pi}{6}$
(c) nπ ± $\frac{n \pi}{4}$
(d) $\frac{n \pi}{2}$
Solution:
(b) $\frac{n \pi}{6}$
[Hint: tan(A + B + C) = $\frac{\tan A+\tan B+\tan C-\tan A \cdot \tan B \cdot \tan C}{1-\tan A \cdot \tan B-\tan B \cdot \tan C-\tan C \cdot \tan A}$
Since , tan θ + tan 2θ + tan 3θ = tan θ ∙ tan 2θ ∙ tan 3θ,
we get, tan (θ + 2θ + 3θ) = θ
∴ tan6θ = 0
∴ 6θ = nπ, θ = $\frac{n \pi}{6}$.]

Question 21.
If any ∆ABC, if a cos B = b cos A, then the triangle is ________.
(a) Equilateral triangle
(b) Isosceles triangle
(c) Scalene
(d) Right angled
Solution:
(b) Isosceles triangle

II: Solve the following
Question 1.
Find the principal solutions of the following equations :
(i) sin2θ = $-\frac{1}{2}$
Solution:
sin2θ = $-\frac{1}{2}$
Since, θ ∈ (0, 2π), 2∈ ∈ (0, 4π)

(ii) tan3θ = -1
Solution:
Since, θ ∈ (0, 2π), 3∈ ∈ (0, 6π)

… [∵ tan(π – θ) = tan(2π – θ) = tan(3π – θ)
= tan (4π – θ) = tan (5π – θ) = tan (6π – θ) = -tan θ]
∴ tan3θ = tan$\frac{3 \pi}{4}$ = tan$\frac{7 \pi}{4}$ = tan$\frac{11 \pi}{4}$ = tan$\frac{15 \pi}{4}$

(iii) cotθ = 0
Solution:
cotθ = 0
Since θ ∈ (0, 2π),
cotθ = 0 = cot $\frac{\pi}{2}$ = cot (π + $\frac{\pi}{2}$ …[∵ cos(π + θ) = cotθ]
∴ cotθ = cot$\frac{\pi}{2}$ = cot$\frac{3 \pi}{2}$
∴ θ = $\frac{\pi}{2}$ or θ = $\frac{3 \pi}{2}$
Hence, the required principal solutions are $\left\{\frac{\pi}{2}, \frac{3 \pi}{2}\right\}$

Question 2.
Find the principal solutions of the following equations :
(i) sin2θ = $-\frac{1}{\sqrt{2}}$
Solution:

(ii) tan5θ = -1
Solution:

(iii) cot2θ = 0
Solution:

Question 3.
Which of the following equations have no solutions ?
(i) cos 2θ = $\frac{1}{3}$
Solution:
cos 2θ = $\frac{1}{3}$
Since $\frac{1}{3}$ ≤ cosθ ≤ 1 for any θ
cos2θ = $\frac{1}{3}$ has solution

(ii) cos² θ = -1
Solution:
cos2θ = -1
This is not possible because cos2θ ≥ 0 for any θ.
∴ cos2θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = $\frac{3}{2}$
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 sin θ = 5
Solution:
3 sin θ = 5
∴ sin θ = $\frac{5}{3}$
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 3 sin θ = 5 does not have any solution.

Question 4.
Find the general solutions of the following equations :
(i) tanθ = $-\sqrt {x}$
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z.
Now, tanθ = $-\sqrt {x}$
∴ tanθ = tan$\frac{\pi}{3}$ …[∵ tan$\frac{\pi}{3}$ = $\sqrt {3}$]
∴ tanθ = tan$\left(\pi-\frac{\pi}{3}\right)$ …[∵ tan(π – θ) = -tanθ]
∴ tanθ = tan$\frac{2 \pi}{3}$
∴ the required general solution is
θ = nπ + $\frac{2 \pi}{3}$, n ∈ Z.

(ii) tan²θ = 3
Solution:
The general solution of tan²θ = tan²∝ is
θ = nπ ± ∝, n ∈ Z.
Now, tan²θ = 3 = ($\sqrt {x}$)²
∴ tan²θ = (tan$\frac{\pi}{3}$)² …[∵ tan$\frac{\pi}{3}$ = $\sqrt {3}$]
∴ tan²θ = tan²$\frac{\pi}{3}$
∴ the required general solution is
θ = nπ ± $\frac{\pi}{3}$, n ∈ Z.

(iii) sin θ – cosθ = 1
Solution:
∴ cosθ – sin θ = -1

(iv) sin²θ – cos²θ = 1
Solution:
sin²θ – cos²θ = 1
∴ cos²θ – sin²θ = -1
∴ cos2θ = cosπ …(1)
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of (1) is given by
2θ = 2nπ ± π, n ∈ Z
∴ θ = nπ ± $\frac{\pi}{2}$, n ∈ Z

Question 5.
In ∆ABC prove that cos $\left(\frac{A-B}{2}\right)=\left(\frac{a+b}{c}\right)$ sin $\frac{C}{2}$
Solution:
By the sine rule,

Question 6.
With usual notations prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^{2}-b^{2}}{c^{2}}$.
Solution:
By the sine rule,
$\frac{a}{\sin \mathrm{A}}$ = $\frac{b}{\sin \mathrm{B}}$ = $\frac{c}{\sin \mathrm{C}}$ = k
∴ a = ksinA, b = ksinB, c = ksinC

Question 7.
In ∆ABC prove that (a – b)² 2cos²$\frac{\mathrm{C}}{2}$ + (a + b)² sin²$\frac{\mathrm{C}}{2}$ = c².
Solution:
LHS (a – b)² 2cos²$\frac{\mathrm{C}}{2}$ + (a + b)² sin²$\frac{\mathrm{C}}{2}$
= (a² + b² – 2ab) cos²$\frac{\mathrm{C}}{2}$ + (a² + b² + 2ab) sin$\frac{\mathrm{C}}{2}$²
= (a² + b²) cos²$\frac{\mathrm{C}}{2}$ – 2ab cos²$\frac{\mathrm{C}}{2}$ + (a² + b²) sin²$\frac{\mathrm{C}}{2}$ + 2ab sin²$\frac{\mathrm{C}}{2}$
= (a² + b²) (cos²$\frac{\mathrm{C}}{2}$ + sin²$\frac{\mathrm{C}}{2}$) – 2ab(cos²$\frac{\mathrm{C}}{2}$ – sin²$\frac{\mathrm{C}}{2}$)
= a² + b² – 2ab cos C
= c² = RHS.

Question 8.
In ∆ABC if cosA = sin B – cos C then show that it is a right angled triangle.
Solution:
cos A= sin B – cos C
∴ cos A + cos C = sin B

∴ A – C = B
∴ A = B + C
∴ A + B + C = 180° gives
A + A = 180°
∴ 2A = 180 ∴ A = 90°
∴ ∆ ABC is a rightangled triangle.

Question 9.
If $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$ then show that a², b², c², are in A.P.
Solution:
By sine rule,
$\frac{\sin \mathrm{A}}{a}$ = $\frac{\sin \mathrm{B}}{b}$ = $\frac{\sin \mathrm{C}}{c}$ = k
∴ sin A = ka, sin B = kb,sin C = kc
Now, $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$
∴ sinA∙sin(B – C) = sinC∙sin(A -B)
∴ sin [π – (B + C)] ∙ sin (B – C)
= sin [π – (A + B)]∙sin (A – B) … [∵ A + B + C = π]
∴ sin(B + C) ∙ sin(B – C) = sin (A + B) ∙ sin (A – B)
∴ sin²B – sin²C = sin²A – sin²B
∴ 2 sin²B = sin²A + sin²C
∴ 2k²b² = k²a² + k²c²
∴ 2b² = a² + c²
Hence, a², b², c² are in A.P.

Question 10.
Solve the triangle in which a = ($\sqrt {3}$ + 1), b = ($\sqrt {3}$ – 1) and ∠C = 60°.
Solution:
Given : a = $\sqrt {3}$ + 1, b = $\sqrt {3}$ – 1 and ∠C = 60°.
By cosine rule,
c² = a² + b² – 2ab cos C
= ($\sqrt {3}$ + 1)² + ($\sqrt {3}$ – 1)² – 2($\sqrt {3}$ + 1)($\sqrt {3}$ – 1)cos60°
= 3 + 1 + 2$\sqrt {3}$ + 3+ 1 – 2$\sqrt {3}$ – 2(3 – 1)$\left(\frac{1}{2}\right)$
= 8 – 2 = 6
∴ c = $\sqrt {6}$ …[∵ c > 0)
By sine rule,

∴ sin A = sin 60° cos 45° + cos 60° sin 45°
and sin B = sin 60° cos 45° – cos 60° sin 45°
∴ sin A = sin (60° + 45°) – sin 105°
and sin B = sin (60° – 45°) = sin 15°
∴ A = 105° and B = 15°
Hence, A = 105°, B 15° and C = $\sqrt {6}$ units.

Question 11.
In ∆ABC prove the following :
(i) a sin A – b sin B = c sin (A – B)
Solution:
By sine rule,
$\frac{a}{\sin \mathrm{A}}$ = $\frac{b}{\sin \mathrm{B}}$ = $\frac{c}{\sin \mathrm{C}}$ = k
∴ a = ksinA, b = ksinB, c = ksinC,
LHS = a sin A – b sinB
= ksinA∙sinA – ksinB∙sinB
= k (sin²A – sin²B)
= k (sin A + sin B)(sin A – sin B)

= k × sin (A + B) × sin (A – B)
= ksin(π – C)∙sin(A – B) … [∵ A + B + C = π]
= k sinC∙sin (A – B)
= c sin (A – B) = RHS.

(ii) $\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}$.
Solution:

(iii) a² sin (B – C) = (b² – c²) sinA
Solution:
By sine rule,
$\frac{a}{\sin \mathrm{A}}$ = $\frac{b}{\sin \mathrm{B}}$ = $\frac{c}{\sin \mathrm{C}}$ = k
∴ a = ksinA, b = ksinB, c = ksinC
RHS = (b² – c²) sin A
= (k²sin²B – k²sin²C)sin A
= k²(sin²B – sin²C) sin A
= k²(sin B + sin C)(sin B – sin C) sin A

= k² × sin (B + C) × sin (B – C) × sin A
= k²∙sin(π – A)∙sin(B – C)∙sinA … [∵ A + B + C = π]
= k²sin A∙sin (B – C)∙sin A
= (k sin A)²∙sin (B – C)
= a²sin (B – C) = LHS.

(iv) ac cos B – bc cos A = (a² – b²).
Solution:
LHS = ac cos B – bc cos A
= ac$\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)$ – bc$\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)$
=$\frac{1}{2}$(c² + a² – b²) – $\frac{1}{2}$(b² + c² – a²)
= $\frac{1}{2}$(c² + a² – b² – b² – c² + a²)
= $\frac{1}{2}$(2a² – 2b²) = a² – b2 = RHS.

(v) $\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{a^{2}+b^{2}+c^{2}}{2 a b c}$ .
Solution:

(vi) $\frac{\cos 2 \mathrm{~A}}{a^{2}}-\frac{\cos 2 \mathrm{~B}}{b^{2}}=\frac{1}{a^{2}}-\frac{1}{b^{2}}$.
Solution:
By sine rule,
$\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}$

(vii) $\frac{b-c}{a}=\frac{\tan \frac{B}{2}-\tan \frac{C}{2}}{\tan \frac{B}{2}+\tan \frac{C}{2}}$
Solution:
By sine rule,

Question 12.
In ∆ABC if a², b², c², are in A.P. then cot$\frac{A}{2}$, cot$\frac{B}{2}$, cot$\frac{C}{2}$ are also in A.P.
Question is modified
In ∆ABC if a, b, c, are in A.P. then cot$\frac{A}{2}$, cot$\frac{B}{2}$, cot$\frac{C}{2}$ are also in A.P.
Solution:
a, b, c, are in A.P.
∴ 2b = a + c …(1)

Question 13.
In ∆ABC if ∠C = 90º then prove that sin(A – B) = $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
Solution:
In ∆ABC, if ∠C = 90º
∴ c² = a² + b² …(1)
By sine rule,

Question 14.
In ∆ABC if $\frac{\cos A}{a}=\frac{\cos B}{b}$, then show that it is an isosceles triangle.
Solution:
Given : $\frac{\cos A}{a}=\frac{\cos B}{b}$ ….(1)
By sine rule,

∴ sin A cos B = cos A sinB
∴ sinA cosB – cosA sinB = 0
∴ sin (A – B) = 0 = sin0
∴ A – B = 0 ∴ A = B
∴ the triangle is an isosceles triangle.

Question 15.
In ∆ABC if sin²A + sin²B = sin²C then prove that the triangle is a right angled triangle.
Question is modified
In ∆ABC if sin²A + sin²B = sin²C then show that the triangle is a right angled triangle.
Solution:
By sine rule,
$\frac{\sin \mathrm{A}}{a}$ = $\frac{\sin \mathrm{B}}{b}$ = $\frac{\sin \mathrm{C}}{c}$ = k
∴ sin A = ka, sinB = kb, sin C = kc
∴ sin²A + sin²B = sin²C
∴ k²a² + k²b² = k²c²
∴ a² + b² = c²
∴ ∆ABC is a rightangled triangle, rightangled at C.

Question 16.
In ∆ABC prove that a²(cos²B – cos²C) + b²(cos²C – cos²A) + c²(cos²A – cos²B) = 0.
Solution:
By sine rule,
$\frac{a}{\sin \mathrm{A}}$ = $\frac{b}{\sin \mathrm{B}}$ = $\frac{c}{\sin \mathrm{C}}$ = k
LHS = a²(cos²B – cos²C) + b²( cos²C – cos²A) + c²(cos²A – cos²B)
= k²sin²A [(1 – sin²B) – (1 – sin²C)] + k²sin²B [(1 – sin²C) – (1 – sin²A)] + k²sin²C[(1 – sin²A) – (1 – sin²B)]
= k²sin²A (sin²C – sin²B) + k²sin²B(sin²A – sin²C) + k²sin²C (sin²B – sin²A)
= k²(sin²A sin²C – sin²Asin²B + sin²A sin²B – sin²B sin²C + sin²B sin²C – sin²A sin²C)
= k²(0) = 0 = RHS.

Question 17.
With usual notations show that (c² – a² + b²) tan A = (a² – b² + c²) tan B = (b² – c² + a²) tan C.
Solution:
By sine rule,
$\frac{a}{\sin A}$ = $\frac{b}{\sin B}$ = $\frac{c}{\sin C}$ = k
∴ a = fksinA, b = ksinB, c = ksinC


From (1), (2) and (3), we get
(c² – a² + b²) tan A = (a² – b² + c²) tan B
= (b² – c² + a²) tan C.

Question 18.
In ∆ABC, if a cos²$\frac{C}{2}$ + c cos²$\frac{A}{2}$ = $\frac{3 b}{2}$, then prove that a , b ,c are in A.P.
Solution:
a cos²$\frac{C}{2}$ + c cos²$\frac{A}{2}$ = $\frac{3 b}{2}$

∴ a + c + b = 3b …[∵ a cos C + c cos A = b]
∴ a + c = 2b
Hence, a, b, c are in A.P.

Question 19.
Show that 2 sin^-1$\left(\frac{3}{5}\right)$ = tan^-1$\left(\frac{24}{7}\right)$.
Solution:
Let sin²$\left(\frac{3}{5}\right)$ = x.



∴ tan^-1$\left(\frac{24}{7}\right)$ = RHS

Question 20.
Show that tan^-1$\left(\frac{1}{5}\right)$ + tan^-1$\left(\frac{1}{7}\right)$ + tan^-1$\left(\frac{1}{3}\right)$ + tan^-1$\left(\frac{1}{8}\right)$ = $\frac{\pi}{4}$.
Solution:

Question 21.
Prove that tan^-1$\sqrt {x}$ = $\frac{1}{2}$ cos^-1$\left(\frac{1-x}{1+x}\right)$, if x ∈ [0, 1].
Solution:
Let tan^-1$\sqrt {x}$ = y
∴ tan y = $\sqrt {x}$ ∴ x = tan²y

Question 22.
Show that $\frac{9 \pi}{8}-\frac{9}{4}$ sin^-1$\frac{1}{3}$ = $\frac{9}{4}$ sin^-1$\frac{2 \sqrt{2}}{3}$.
Question is modified
Show that $\frac{9 \pi}{8}-\frac{9}{4}$ sin^-1$\left(\frac{1}{3}\right)$ = $\frac{9}{4}$ sin^-1$\left(\frac{2 \sqrt{2}}{3}\right)$.
Solution:
We have to show that

Question 23.
Show that

Solution:

Question 24.
If sin(sin^-1$\frac{1}{5}$ + cos^-1x) = 1, then find the value of x.
Solution:
sin(sin^-1$\frac{1}{5}$ = 1

Question 25.
If tan^-1$\left(\frac{x-1}{x-2}\right)$ + tan^-1$\left(\frac{x+1}{x+2}\right)$ = $\frac{\pi}{4}$ then find the value of x.
Solution:
tan^-1$\left(\frac{x-1}{x-2}\right)$ + tan^-1$\left(\frac{x+1}{x+2}\right)$ = $\frac{\pi}{4}$

∴ x = ±$\frac{1}{\sqrt{2}}$.

Question 26.
If 2 tan^-1(cos x ) = tan^-1(cosec x) then find the value of x.
Solution:
2 tan^-1(cos x ) = tan^-1(cosec x)

Question 27.
Solve: tan^-1$\left(\frac{1-x}{1+x}\right)$ = $\frac{1}{2}$(tan^-1x), for x > 0.
Solution:
tan^-1$\left(\frac{1-x}{1+x}\right)$ = $\frac{1}{2}$(tan^-1x)

Question 28.
If sin^-1(1 – x) – 2sin^-1x = $\frac{\pi}{2}$, then find the value of x.
Solution:
sin^-1(1 – x) – 2sin^-1x = $\frac{\pi}{2}$

Question 29.
If tan^-12x + tan^-13x = $\frac{\pi}{4}$, then find the value of x.
Question is modified
If tan^-12x + tan^-13x = $\frac{\pi}{2}$, then find the value of x.
Solution:
tan^-12x + tan^-13x = $\frac{\pi}{4}$
∴ tan^-1$\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)$ = tan$\frac{\pi}{4}$, where 2x > 0, 3x > 0
∴ $\frac{5 x}{1-6 x^{2}}$ = tan$\frac{\pi}{4}$ = 1
∴ 5x = 1 – 6x²
∴ 6x² + 5x – 1 = 0
∴ 6x² + 6x – x – 1 = 0
∴ 6x(x +1) – 1(x + 1) = 0
∴ (x + 1)(6x – 1) = 0
∴ x = -1 or x = $\frac{1}{6}$
But x > 0 ∴ x ≠ -1
Hence, x = $\frac{1}{6}$

Question 30.
Show that tan^-1$\frac{1}{2}$ – tan^-1$\frac{1}{4}$ = tan^-1$\frac{2}{9}$.
Solution:
LHS = tan^-1$\frac{1}{2}$ – tan^-1$\frac{1}{4}$

Question 31.
Show that cot^-1$\frac{1}{3}$ – tan^-1$\frac{1}{3}$ = cot^-1$\frac{3}{4}$.
Solution:
LHS = cot^-1$\frac{1}{3}$ – tan^-1$\frac{1}{3}$

Question 32.
Show that tan^-1$\frac{1}{2}$ = $\frac{1}{3}$ tan^-1$\frac{11}{2}$.
Solution:
We have to show that

Question 33.
Show that cos^-1$\frac{\sqrt{3}}{2}$ + 2sin^-1$\frac{\sqrt{3}}{2}$ = $\frac{5 \pi}{6}$
Solution:

Question 34.
Show that 2cot^-1$\frac{3}{2}$ + sec^-1$\frac{13}{12}$ = $\frac{\pi}{2}$
Solution:

Question 35.
Prove the following :
(i) cos^-1 x = tan^-1$\frac{\sqrt{1-x^{2}}}{x}$, if x < 0.
Question is modified
cos^-1 x = tan^-1$\left(\frac{\sqrt{1-x^{2}}}{x}\right)$, if x > 0.
Solution:

(ii) cos^-1 x = π + tan^-1$\frac{\sqrt{1-x^{2}}}{x}$, if x < 0.
Solution:

Question 36.
If |x| < 1 , then prove that 2tan^-1 x = tan^-1$\frac{2 x}{1-x^{2}}$ = sin^-1$\frac{2 x}{1+x^{2}}$ = cos^-1$\frac{1-x^{2}}{1+x^{2}}$
Question is modified
If |x| < 1 , then prove that 2tan^-1 x = tan^-1$\left(\frac{2 x}{1-x^{2}}\right)$ = sin^-1$\left(\frac{2 x}{1+x^{2}}\right)$ = cos^-1$\left(\frac{1-x^{2}}{1+x^{2}}\right)$
Solution:
Let tan^-1x = y
Then, x = tany

Question 37.
If x, y, z, are positive then prove that tan^-1$\frac{x-y}{1+x y}$ + tan^-1$\frac{y-z}{1+y z}$ + tan^-1$\frac{z-x}{1+z x}$ = 0
Solution:

Question 38.
If tan^-1 x + tan^-1 y + tan^-1 z = $\frac{\pi}{2}$ then, show that xy + yz + zx = 1
Solution:
tan^-1 x + tan^-1 y + tan^-1 z = $\frac{\pi}{2}$

∴ 1 – xy – yz – zx = 0
∴ xy + yz + zx = 1.

Question 39.
If cos^-1 x + cos^-1 y + cos^-1 z = π then show that x² + y² + z² + 2xyz = 1.
Solution:
0 ≤ cos^-1x ≤ π and
cos^-1x + cos^-1y+ cos^-1z = 3π
∴ cos^-1x = π, cos^-1y = π and cos^-1z = π
∴ x = y = z = cosπ = -1
∴ x² + y² + z² + 2xyz
= (-1)² + (-1)² + (-1)² + 2(-1)(-1)(-1)
= 1 + 1 + 1 – 2
= 3 – 2 = 1.