## Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

### Introduction to Trigonometry Class 10 Important Questions Very Short Answer (1 Mark)

Trigonometry Questions For Class 10 Chapter 8 Question 1.
If tan Î¸ + cot Î¸ = 5, find the value of tan2Î¸ + cotÎ¸. (2012)
Solution:
tan Î¸ + cot Î¸ = 5 … [Given
tan2Î¸ + cot2Î¸ + 2 tan Î¸ cot Î¸ = 25 … [Squaring both sides
tan2Î¸ + cot2Î¸ + 2 = 25
âˆ´ tan2Î¸ + cot2Î¸ = 23

Trigonometry Important Questions Class 10 Chapter 8 Question 2.
If sec 2A = cosec (A – 27Â°) where 2A is an acute angle, find the measure of âˆ A. (2012, 2017D)
Solution:
sec 2A = cosec (A – 27Â°)
cosec(90Â° – 2A) = cosec(A – 27Â°) …[âˆµ sec Î¸ = cosec (90Â° – Î¸)
90Â° – 2A = A – 27Â°
90Â° + 27Â° = 2A + A
â‡’ 3A = 117Â°
âˆ´ âˆ A = $$\frac{117^{\circ}}{3}$$ = 39Â°

Class 10 Trigonometry Questions Chapter 8 Question 3.
If tan Î± = $$\sqrt{3}$$ and tan Î² = $$\frac{1}{\sqrt{3}}$$,0 < Î±, Î² < 90Â°, find the value of cot (Î± + Î²). (2012)
Solution:
tan Î± = $$\sqrt{3}$$ = tan 60Â° …(i)
tan Î² = $$\frac{1}{\sqrt{3}}$$ = tan 30Â° …(ii)
Solving (i) & (ii), Î± = 60Â° and Î² = 30Â°
âˆ´ cot (Î± + Î²) = cot (60Â° + 30Â°) = cot 90Â° = 0

Important Questions For Class 10 Maths Trigonometry With Solutions Question 4.
If sin Î¸ – cos Î¸ = 0, find the value of sin4 Î¸ + cos4 Î¸. (2012, 2017D)
Solution:
sin Î¸ – cos Î¸ = 0 = sin Î¸ = cos Î¸
â‡’ $$\frac{\sin \theta}{\cos \theta}$$ = 1 â‡’ tan Î¸ = 1 â‡’ Î¸ = 45Â°
Now, sin4Î¸ + cos4Î¸
= sin4 45Â° + cos4 45Â°
= $$\left(\frac{1}{\sqrt{2}}\right)^{4}+\left(\frac{1}{\sqrt{2}}\right)^{4}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$

Trigonometry Class 10 Important Questions Chapter 8 Question 5.
If sec Î¸ + tan Î¸ = 7, then evaluate sec Î¸ – tan Î¸. (2017OD)
Solution:
We know that,
sec2Î¸ – tan2Î¸ = 1
(sec Î¸ + tan Î¸) (sec Î¸ – tan Î¸) = 1
(7) (sec Î¸ – tan Î¸) = 1 …[sec Î¸ + tan Î¸ = 7; (Given)
âˆ´ sec Î¸ – tan Î¸ = $$\frac{1}{7}$$

Questions On Trigonometry Class 10 Chapter 8 Question 6.
Evaluate: 10. $$\frac{1-\cot ^{2} 45^{\circ}}{1+\sin ^{2} 90^{\circ}}$$. (2014)
Solution:

Trigonometry Questions Class 10 Chapter 8 Question 7.
If cosec Î¸ = $$\frac{5}{4}$$, find the value of cot Î¸. (2014)
Solution:
We know that, cot2Î¸ = cosec2Î¸ – 1
= $$\left(\frac{5}{4}\right)^{2}$$ – 1 â‡’ $$\frac{25}{16}$$ – 1 â‡’ $$\frac{25-16}{16}$$
coÅ¥2Î¸ = $$\frac{9}{16}$$ i cot Î¸ = $$\frac{3}{4}$$

Trigonometry Class 10 Questions Chapter 8 Question 8.
If Î¸ = 45Â°, then what is the value of 2 sec2Î¸ + 3 cosec2Î¸ ? (2014)
Solution:
2 sec2Î¸ + 3 cosec2Î¸ = 2 sec2 45Â° + 3 cosec2 45Â°
= 2($$\sqrt{2}$$)2 + 3 ($$\sqrt{2}$$)2 = 4 + 6 = 10

10th Trigonometry Questions Chapter 8 Question 9.
If $$\sqrt{3}$$ sin Î¸ = cos Î¸, find the value of $$\frac{3 \cos ^{2} \theta+2 \cos \theta}{3 \cos \theta+2}$$. (2015)
Solution:
$$\sqrt{3}$$ sin Î¸ = cos Î¸ … [Given

Important Questions For Class 10 Maths Trigonometry Pdf Question 10.
Evaluate: sin2 19Â° + sin771Â°. (2015)
Solution:
sin2 19Â° + sin2 71Â°
= sin219Â° + sin2 (90Â° – 19Â°)…[âˆµ sin(90Â° – Î¸) = cos Î¸
= sin2 19Â° + cos2 19Â° = 1 …[âˆµ sin2 Î¸ + cos2 Î¸ = 1

Important Questions Of Trigonometry Class 10 With Solutions PdfÂ  Question 11.
What happens to value of cos when increases from 0Â° to 90Â°? (2015)
Solution:
cos 0Â° = 1, cos 90Â° = 0
When Î¸ increases from 0Â° to 90Â°, the value of cos Î¸ decreases from 1 to 0.

Class 10th Trigonometry Questions Chapter 8 Question 12.
If tan Î¸ = $$\frac{a}{x}$$, find the value of $$\frac{x}{\sqrt{a^{2}+x^{2}}}$$. (2013)
Solution:

Trigonometry Questions For Class 10 With Solutions Question 13.
If in a right angled âˆ†ABC, tan B = $$\frac{12}{5}$$, then find sin B. (2014)
Solution:
1st method:
tan B = $$\frac{12}{5}$$ âˆ´ cot B = $$\frac{5}{12}$$
cosec2 B = 1 + cot2 B
= 1 + $$[\left(\frac{5}{12}\right)^{2}/latex] = 1 + [latex]$$
= $$\frac{144+25}{144}=\frac{169}{144}$$
cosec B = $$\frac{13}{12}$$ âˆ´ sin B = $$\frac{12}{13}$$
2nd method:

tan B = $$\frac{12}{5}$$
tan B = $$\frac{AC}{BC}$$
Let AC = 12k, BC = 5k
In rt. âˆ†ACB,
AB2 = AC2 + BC2 …[Pythagoras theorem
AB2 = (12k)2 + (5k)2
AB2 = 144k2 + 25k22 = 169k2
AB = 13k
âˆ´ sin B = $$\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{12 \mathrm{k}}{13 \mathrm{k}}=\frac{12}{13}$$

Important Trigonometry Questions For Class 10 Question 14.
If âˆ†ABC is right angled at B, what is the value of sin (A + C). (2015)
Solution:

âˆ B = 90Â° …[Given
âˆ A + âˆ B + âˆ C = 180Â° …[Angle sum property of a âˆ†
âˆ A + âˆ C + 90Â° = 180Â°
âˆ A + âˆ C = 90Â°
âˆ´ sin (A + C) = sin 90Â° = 1 …(taking sin both side

### Introduction to Trigonometry Class 10 Important Questions Short Answer-I (2 Marks)

Important Question Of Trigonometry Class 10 Question 15.
Evaluate: tan 15Â° . tan 25Â° , tan 60Â° . tan 65Â° . tan 75Â° – tan 30Â°. (2013)
Solution:
tan 15Â°. tan 25Â°, tan 60Â°. tan 65Â°. tan 75Â° – tan 30Â°
= tan(90Â° – 75Â°) tan(90Â° – 65Â°). $$\sqrt{3}$$ . tan 65Â°. tan 75Â° – $$\frac{1}{\sqrt{3}}$$

Question 16.
Express cot 75Â° + cosec 75Â° in terms of trigonometric ratios of angles between 0Â° and 30Â°. (2013)
Solution:
cot 75Â° + cosec 75Â°
= cot(90Â° – 15Â°) + cosec(90Â° – 15Â°)
= tan 15Â° + sec 15Â° …[cot(90Â°-A) = tan A
cosec(90Â° – A) = sec A

Question 17.
If cos (A + B) = 0 and sin (A – B) = 3, then find the value of A and B where A and B are acute angles. (2012)
Solution:

Putting the value of B in (i), we get
â‡’ A = 30Â° + 30Â° = 60Â°
âˆ´ A = 60Â°, B = 30Â°

Question 18.
If A, B and C are the interior angles of a âˆ†ABC, show that sin $$\left(\frac{A+B}{2}\right)$$ = cos$$\left(\frac{c}{2}\right)$$. (2012)
Solution:
In âˆ†ABC, âˆ A + âˆ B + âˆ C = 180Â° …(Angle sum property of âˆ†
âˆ A + âˆ B = 180Â° – âˆ C

Question 19.
If x = p sec Î¸ + q tan Î¸ and y = p tan Î¸ + q sec Î¸, then prove that x2 – y2 = p2 – q2. (2014)
Solution:
L.H.S. = x2 – y2
= (p sec Î¸ + q tan Î¸)2 â€“ (p tan Î¸ + q sec Î¸)2
= p2 sec Î¸ + q2 tan2 Î¸ + 2 pq sec 2 tan 2 -(p2 tan2 Î¸ + q2 sec2 Î¸ + 2pq sec Î¸ tan Î¸)
= p2 sec Î¸ + 2 tan2 Î¸ + 2pq sec Î¸ tan Î¸ – p2 tan2 Î¸ – q2 sec Î¸ – 2pq sec Î¸ tan Î¸
= p2(sec2 Î¸ â€“ tan2 Î¸) – q2(sec?2 Î¸ – tan2 Î¸) =
= p2 – q2 …[sec2 Î¸ – tan2 Î¸ = 1
= R.H.S.

Question 20.
Prove the following identity: (2015)
$$\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}$$ = 1 – sin Î¸ . cos Î¸
Solution:

Question 21.
Simplify: $$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$$. (2014)
Solution:

Question 22.
If x = a cos Î¸ – b sin Î¸ and y = a sin Î¸ + b cos Î¸, then prove that a2 + b2 = x2 + y2. (2015)
Solution:
R.H.S. = x2 + y2
= (a cos Î¸ – b sin Î¸)2 + (a sin Î¸ + b cos Î¸)2
= a2cos2 Î¸ + b2 sin2 Î¸ – 2ab cos Î¸ sin Î¸ + a2 sin2 Î¸ + b2 cos2 Î¸ + 2ab sin Î¸ cos Î¸
= a2(cos2 Î¸ + sin2Î¸) + b2 (sin2 Î¸ + cos2 Î¸)
= a2 + b2 = L.H.S. …[âˆµ cos2 Î¸ + sin2 Î¸ = 1

### Introduction to Trigonometry Class 10 Important Questions Short Answer – II (3 Marks)

Question 23.
Given 2 cos 3Î¸ = $$\sqrt{3}$$, find the value of Î¸. (2014)
Solution:
2 cos 3Î¸ = $$\sqrt{3}$$ …[Given
cos 3Î¸ = $$\frac{\sqrt{3}}{2}$$ â‡’ cos 3Î¸ = cos 30Â°
30 = 30Â° âˆ´ Î¸ = 10Â°

Question 24.
If cos x = cos 40Â° . sin 50Â° + sin 40Â°. cos 50Â°, then find the value of x. (2014)
Solution:
cos x = cos 40Â° sin 50Â° + sin 40Â° cos 50Â°
cos x = cos 40Â° sin(90Â° – 40Â°) + sin 40Â°.cos(90Â° – 40Â°)
cos x = cos2 40Â° + sin2 40Â°
cos x = 1 …[âˆµ cos2 A + sin2 A = 1
cos x = cos 0Â° â‡’ x = 0Â°

Question 25.
If sin Î¸ = $$\frac{1}{2}$$, then show that 3 cos Î¸ – 4 cos3 Î¸ = 0. (2014)
Solution:
sin Î¸ = $$\frac{1}{2}$$
sin Î¸ = sin 30Â° â‡’ Î¸ = 30Â°
L.H.S = 3 cos Î¸ – 4 cos3 Î¸
= 3 cos 30Â° – 4 cos3(30Â°)

Question 26.
If 5 sin Î¸ = 4, prove that $$\frac{1}{\cos \theta}+\frac{1}{\cot \theta}$$ = 3 (2013
Solution:
Given: 5 sin Î¸ = 4

Question 27.
Evaluate: sec 41Â°. sin 49Â° + cos 29Â°.cosec 61Â° Â (2012)
Solution:

Question 28.
Evaluate: (2012, 2017D)

Solution:

Question 29.
In figure, âˆ†PQR right angled at Q, PQ = 6 cm and PR = 12 cm. Determine âˆ QPR and âˆ PRQ. (2013)

Solution:
In rt. âˆ†PQR,
PQ2 + QR2 = PR2 …[By Pythogoras’ theorem

(6)2 + QR2 = (12)2
QR2 = 144 – 36
QR2 = 108

Question 30.
Find the value of: (2013)

Solution:

Question 31.
Prove that: $$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\sec ^{2} 20^{\circ}-\cot ^{2} 70^{\circ}}$$ + 2 sin 36Â° sin 42Â° sec 48Â° sec 54Â° (2017OD)
Solution:

Question 32.
If sin Î¸ = $$\frac{12}{13}$$, 0Â° <0 < 90Â°, find the value of: $$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cdot \cos \theta} \times \frac{1}{\tan ^{2} \theta}$$ (2015)
Solution:

Question 33.
Prove that: (2012)

Solution:

Question 34.
Prove that: $$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{1+\sin \theta}{\cos \theta}$$ (2012, 2017D)
Solution:

Question 35.
If tan Î¸ = $$\frac{a}{b}$$, prove that $$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$$ (2013)
Solution:

Question 36.
Prove the identity: (sec A – cos A). (cot A + tan A) = tan A . sec A. (2014)
Solution:
L.H.S.= (sec A – cos A) (cot A + tan A)

Question 37.
If sec Î¸ + tan Î¸ = p, prove that sin Î¸ = $$\frac{p^{2}-1}{p^{2}+1}$$ (2015)
Solution:

Question 38.
Prove that: $$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}$$ = tan Î¸ (2015)
Solution:

Question 39.
Prove that: $$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$$ = 2 cosec Î¸ (2017OD)
Solution:

### Introduction to Trigonometry Class 10 Important Questions Long Answer (4 Marks)

Question 40.
In an acute angled triangle ABC, if sin (A + B – C) = $$\frac{1}{2}$$ and cos (B + C – A) = $$\frac{1}{\sqrt{2}}$$, find âˆ A, âˆ B and âˆ C. (2012)
Solution:

Putting the values of A and B in (iii), we get
67.5Â° + B + 75o = 180Â°
B = 180Â° – 67.5Â° – 75o = 37.5Â°
âˆ´ âˆ A = 67.5Â°, âˆ B = 37.5Â° and âˆ C = 75Â°

Question 41.
Evaluate: (2013)

Solution:

Question 42.
Evaluate the following: (2015)

Solution:

Question 43.
If Î¸ = 30Â°, verify the following: (2014)
(i) cos 3Î¸ = 4 cos3 Î¸ – 3 cos Î¸
(ii) sin 3Î¸ = 3 sin Î¸ – 4 sin3Î¸
Solution:

Question 44.
If tan (A + B) = $$\sqrt{3}$$ and tan (A – B) = $$\frac{1}{\sqrt{3}}$$ where 0 < A + B < 90Â°, A > B, find A and B. Also calculate: tan A. sin (A + B) + cos A. tan (A – B). (2015)
Solution:

Question 45.
Find the value of cos 60Â° geometrically. Hence find cosec 60Â°. (2012, 2017D)
Solution:

Let âˆ†ABC be an equilateral âˆ†.
Let each side of triangle be 2a.
Since each angle in an equilateral âˆ† is 60Â°
âˆ´ âˆ A = âˆ B = âˆ C = 60Â°
AB = AC … [Each = 2a
âˆ 1 -âˆ 2 … [Each 90Â°
BD = DC = $$\frac{2 a}{2}$$ = a
In rt. âˆ†ADB, cos 60Â° = $$\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{a}{2 a}=\frac{1}{2}$$

Question 46.
If tan(20Â° – 3Î±) = cot(5Î± – 20Â°), then find the value of Î± and hence evaluate: sin Î±. sec Î± . tan Î± – cosec Î± . cos Î± . cot Î±. (2014)
Solution:
tan(20Â° – 3Î±) = cot(5Î± – 20Â°)
tan(20Â° â€“ 3Î±) = tan[90Â° â€“ (5Î± – 20Â°)] …[âˆµ cot Î¸ = tan(90Â° – Î¸)]
âˆ´ 20Â° – 3Î± = 90Â° – 5Î± + 20Â°
â‡’ -3Î± + 5Î± = 90Â° + 20Â° – 20Â°
â‡’ 2Î± = 90Â° â‡’ Î± = 45Â°
Now, sin Î± . sec Î± tan Î± – cosec Î± . cos Î± . cot Î±
= sin 45Â°. sec 45Â° tan 45Â° – cosec 45Â°. cos 45Â° cot 45Â°
= $$\frac{1}{\sqrt{2}} \times \sqrt{2} \times 1-\sqrt{2} \times \frac{1}{\sqrt{2}} \times 1=1-1=0$$

Question 47.
If $$\frac{x}{a}$$cosÎ¸ + $$\frac{y}{b}$$sinÎ¸ = 1 and $$\frac{x}{a}$$sinÎ¸ – $$\frac{y}{b}$$ cosÎ¸ = 1, prove that event $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 2. (2012, 2017D)
Solution:

Question 48.
If sin Î¸ = $$\frac{c}{\sqrt{c^{2}+d^{2}}}$$ and d > 0, find the values of cos Î¸ and tan Î¸. (2013)
Solution:

Question 49.
If cot B = $$\frac{12}{5}$$, prove that tan2B – sin2B = sin4 B . sec2 B. (2013)
Solution:
cot B = $$\frac{12}{5}$$ :: $$\frac{A B}{B C}=\frac{12}{5}$$

AB = 12k, BC = 5k
In rt. âˆ†ABC, …[By Pythagoras’ theorem
AC2 = AB2 + BC2
AC2 = (12k)2 + (5k)2
AC2 = 144k2 + 25k2
AC2 = 169k2
AC = +13k …[âˆµ Hypotenuse cannot be -ve

Question 50.
If $$\sqrt{3}$$ cot2Î¸ – 4 cot Î¸ + $$\sqrt{3}$$ = 0, then find the value of cot2 Î¸ + tan2Î¸. (2013)
Solution:

Question 51.
Prove that b2x2 – a2y2 = a2b2, if: (2014)
(i) x = a sec Î¸, y = b tan Î¸
(ii) x = a cosec Î¸, y = b cot Î¸
Solution:
(i) L.H.S. = b2x2 – a2y2
= b2(a sec Î¸)2 – a2(b tan Î¸)2
= b2a2 sec Î¸ – a2b2 tan2Î¸
= b2a2(sec2 Î¸ – tan2 Î¸)
= b2a2(1) …[âˆµ sec2Î¸ – tan2 Î¸ = 1
= a2b2 = R.H.S.

(ii) L.H.S. = b2x2 – a2y2
= b2(a cosec Î¸)2 – a2(b cot Î¸)2
= b2a2 cosec2 Î¸ – a2b2 cot2 Î¸
= b2a2(cosec2Î¸ – cot2 Î¸)
= b2a2 (1) ..[âˆµ cosec2 Î¸ – cot2 Î¸ = 1
= a2b2= R.H.S.

Question 52.
If sec Î¸ – tan Î¸ = x, show that sec Î¸ + tan Î¸ = $$\frac{1}{x}$$ and hence find the values of cos Î¸ and sin Î¸. (2015)
Solution:

Question 53.
If cosec Î¸ + cot Î¸ = p, then prove that cos Î¸ = $$\frac{p^{2}-1}{p^{2}+1}$$. (2012)
Solution:
cosec Î¸ + cot Î¸ = p

Question 54.
If tan Î¸ + sin Î¸ = p; tan Î¸ – sin Î¸ = q; prove that p2 – q2 = $$4 \sqrt{p q}$$. (2012)
Solution:
L.H.S. = p2 – q2
= (tan Î¸ + sin Î¸)2 â€“ (tan Î¸ – sin Î¸)2
= (tan2Î¸ + sin2Î¸ + 2.tanÎ¸.sinÎ¸) – (tan2Î¸ + sin2Î¸ – 2tan Î¸ sin Î¸)
= 2 tan Î¸ sin Î¸+ 2 tan Î¸ sin Î¸
= 4 tan Î¸ sin Î¸ …(i)

Question 55.
If sin Î¸ + cos Î¸ = m and sec Î¸ + cosec Î¸ = n, then prove that n(m2 – 1) = 2m. (2013)
Solution:
m2 – 1 = (sin Î¸ + cos Î¸)2 – 1
= sin2 Î¸ + cos2Î¸ + 2 sin Î¸ cos Î¸ – 1
= 1 + 2 sin Î¸ cos Î¸ – 1
= 2 sin Î¸ cos Î¸ …[sin2 Î¸ + cos2 Î¸ = 1
L.H.S. = n(m2 – 1)
= (sec Î¸ + cosec Î¸) 2 sin Î¸ cos Î¸

Question 56.
Prove that: Â = 2 cosec A (2012)
Solution:

Question 57.
In âˆ†ABC, show that sin2 $$\frac{\mathbf{A}}{2}$$ + sin2 $$\frac{\mathbf{B}+\mathbf{C}}{\mathbf{2}}$$ = 1. (2013)
Solution:
In âˆ†ABC, âˆ A + âˆ B + âˆ C = 180Â° … [Sum of the angles of âˆ†
âˆ B + âˆ C = 180Â° – âˆ A

Question 58.
Find the value of: (2013)

Solution:

Question 59.
Prove that: (sin Î¸ + cos Î¸ + 1). (sin Î¸ – 1 + cos Î¸) . sec Î¸ . cosec Î¸ = 2 (2014)
Solution:
L.H.S. = (sin Î¸ + cos Î¸ + 1) (sin Î¸ – 1 + cos Î¸) . sec Î¸ cosec Î¸
= [(sin Î¸ + cos Î¸) + 1] [(sin Î¸ + cos Î¸) – 1] . sec Î¸ cosec Î¸
= [(sin Î¸ + cos Î¸)2 â€“ (1)2] sec Î¸ cosec Î¸ …[âˆµ (a + b)(a – b) = a2 – b2
= (sin2 Î¸ + cos2Î¸ + 2 sin Î¸ cos Î¸ – 1]. sec Î¸ cosec Î¸
= (1 + 2 sin Î¸ cos Î¸ – 1). sec Î¸ cosecÎ¸ …[âˆµ sin2Î¸ + cos2Î¸ = 1
= (2 sin Î¸ cos Î¸). $$\frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}$$
= 2 = R.H.S. …(Hence proved)

Question 60.
Prove that: (2014)

Solution:

Question 61.
Prove that: (1 + cot A + tan A). (sin A – cos A) = $$\frac{\sec ^{3} A-\csc ^{3} A}{\sec ^{2} A \cdot \csc ^{2} A}$$ (2015)
Solution:

Question 62.
Prove the identity: (2015)

Solution:

Question 63.
Prove the following trigonometric identities: sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A. (2015)
Solution:
L.H.S.
= sin A (1 + tan A) + cos A (1 + cot A)

Question 64.
Prove that: (cot A + sec B)2 â€“ (tan B – cosec A)2 = 2(cot A . sec B + tan B. cosec A) (2014)
Solution:
L.H.S.
= (cot A + sec B)2 â€“ (tan B – cosec A)2
= cot2 A + sec2 B + 2 cot A sec B – (tan2 B + cosec2 A – 2 tan B cosec A)
= cot2 A + sec2 B + 2 cot A sec B – tan2 B – cosec2 A + 2 tan B cosec A
= (sec2 B – tan2 B) – (cosec2 A – cot2 A) + 2(cot A sec B + tan B cosec A)
= 1 – 1 + 2(cot A sec B + tan B cosec A) … [âˆµ sec2B – tan2 B = 1
cosec2A – cot2 A = 1
= 2(cot A . sec B + tan B . cosec A) = R.H.S.

Question 65.
If x = r sin A cos C, y = r sin A sin C and z = r cos A, then prove that x2 + y2 + z2 = r2. (2017OD)
Solution:
x = r sin A cos C; y = r sin A sin C; z = r cos A
L.H.S. x2 + y2 + z2 = 2 sin2 A cos2C + r2 sin2 A sin2 C + r2 cos2 A
= r2 sin2 A(cos2 C + sin2 C) + r2 cos2 A
= r2 sin2 A + r2 cos2 A … [cos2Î¸ + sin2Î¸ = 1
= r2 (sin2 A + cos2 A) = r2 = R.H.S.

Question 66.
Prove that: (2017OD)

Solution:

Question 67.
In the adjoining figure, ABCD is a rectanlge with breadth BC = 7 cm and âˆ CAB = 30Â°. Find the length of side AB of the rectangle and length of diagonal AC. If the âˆ CAB = 60Â°, then what is the size of the side AB of the rectangle. [Use $$\sqrt{3}$$ = 1.73 and $$\sqrt{2}$$ = 1.41, if required) (2014OD)

Solution: