TS 10th Class Maths Important Questions Chapter 11 Trigonometry

These TS 10th Class Maths Chapter Wise Important Questions Chapter 11 Trigonometry given here will help you to solve different types of questions.

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Previous Exams Questions

Question 1.
If sin A = cos A then find the value of A. (A.P. Mar. ’15)
Solution:
sin A = cos A (given)
then A + A = 90 ⇒ 2A = 90
⇒ A = \(\frac{90}{2}\) = 45°
∴ A = 45°

Question 2.
If 4 sin2 θ – 1 = 0 then find ‘θ’ (θ < 90) also, find the value of θ and the value of cos2θ + tan2θ (AP. Mar. ’15)
Solution:
Given, 4 Sin2θ – 1 = 0
4 Sin2θ = 1
Sin2θ = \(\frac{1}{4}\)
Sinθ = ± \(\sqrt{\frac{1}{4}}\) = ±\(\frac{1}{2}\)
Given θ is less than 90°
∴ Sin θ = \(\frac{1}{2}\)
sinθ = sin 30°
∴ θ = 30°
Cosθ = Cos 30° = \(\frac{\sqrt{3}}{2}\)
Tan θ = Tan 30° = \(\frac{1}{\sqrt{3}}\)
Cos2 θ + Tan2 θ = Cos2 30° + Tan2 30°
= \(\left(\frac{\sqrt{3}}{2}\right)^2\) + \(\left(\frac{1}{\sqrt{3}}\right)^2\)
= \(\frac{3}{4}\) + \(\frac{1}{3}\) = \(\frac{9+4}{12}\) = \(\frac{13}{12}\)

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 3.
Show that tan2θ – \(\frac{1}{\cos ^2 \theta}\) = 1 (T.S. Mar. ’15)
Solution:
Method – I : Since \(\frac{1}{\cos ^2 \theta}\) = sec2θ
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 12

Question 4.
Find the value of (T.S. Mar. ’15)
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 13
Solution:
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 14

Question 5.
If tan θ = \(\sqrt{3}\) (θ is acute angle) then find the value of 1 + cos θ. (T.S. Mar. ’16)
Solution:
tan θ = \(\sqrt{3}\) = tan 60 (∵ θ is acute)
⇒ θ = 60
⇒ 1 + cos θ = 1 + cos 60 = 1 + \(\frac{1}{2}\) = \(\frac{3}{2}\)
∴ 1 + cos θ = \(\frac{3}{2}\)

Question 6.
Show that \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = sec θ – tan θ. (T.S. Mar. ’16)
Solution:
\(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = \(\sqrt{\frac{(1-\sin \theta)(1-\sin \theta)}{(1+\sin \theta)(1-\sin \theta)}}\)
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 15
Hence proved

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 7.
If tan (A + B) = 1, and cos (A – B) = \(\frac{\sqrt{3}}{2}\) 0° < A + B < 90, A > B then find values of A and B. (T.S. Mar. ’16)
Solution:
tan (A + B) = 1 = tan 45°
∴ A + B = 45° ……………. (1)
cos (A – B) = \(\frac{\sqrt{3}}{2}\) = cos 30°
⇒ A – B = 30° ……………… (2)
Solving the equation (1) and (2) we get
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 16
then A + B = 45
37.5 + B = 45 ⇒ B = 45 – 37.5 = 7.5
So, A = 37.5°, B = 7.5°

Question 8.
Find the value of tan2 60 + 4 cos2 45 + 3 sec2 30 + 5 cos2 90 = cosec 30 + sec 60 – cot2 30 (T.S. Mar. ’16)
Solution:
Put the following values in the given problem
tan 60° = \(\sqrt{3}\) , cos 45° = \(\sqrt{2}\), sec 30° = \(\frac{2}{\sqrt{3}}\)
cos 90° = 0, cosec 30° = 2, sec 60° = 2, cot 30° = \(\sqrt{3}\)
We get
tan2 60 + 4 cos2 45 + 3 sec2 30 + 5 cos2 90 = cosec 30 + sec 60 – cot2 30
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 17

Additional Questions

Question 1.
In a right angled triangle ABC, with right angle at B in which a = 5 units, b = 13 units and ∠BCA = θ, then find sin θ and tan θ.
Solution:
Given a = 5 units = BC
b = 13 units = CA or AC
∠BCA = θ
By pythagoras theorem
AC2 = AB2 + BC2
⇒ 132 = AB2 + 52
⇒ AB2 = 132 – 52 = 169 – 25 = 144
⇒ AB = \(\sqrt{144}\) = 12 units
Now from the figure
Sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\)
and Tan θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{5}\)

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 2.
If Cos c = \(\frac{3}{5}\), then find Sin c and Tan c
Solution:
We have cos c = \(\frac{3}{5}\) = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 1
⇒ AC = 5, BC = 3
By Pythagoras theorem
AC2 = AB2 + BC2
⇒ 52 = AB2 + 32
⇒ 25 = AB2 + 9 = 16
⇒ AB = \(\sqrt{16}\) = 4
From Sin c = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{4}{5}\), Tan c = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{4}{3}\)

Question 3.
If 12 Tan A = 9, then find Sin A and Cos A.
Solution:
Given 12 Tan A = 9
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 2
⇒ Tan A = \(\frac{9}{12}\) = \(\frac{\mathrm{BC}}{\mathrm{AB}}\)
⇒ BC = 9, AB = 12
By Pythagoras theorem
AC2 = AB2 + BC2
= 122 + 92
= 144 + 81
AC2 = 225
AC = \(\sqrt{225}\) = 15
Now, sin A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\) = \(\frac{9}{5}\) and cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{15}\)
∴ sin A = \(\frac{9}{15}\) and cos A = \(\frac{12}{15}\)

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 4.
If 5Cot A = 12, find Cos A and Cosec A.
Solution:
Given Cot A = \(\frac{12}{5}\) = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
⇒ AB = 12, BC = 5
By Pythagoras theorem
AC2 = AB2 + BC2
= 122 + 52
= 144 + 25
= AC2 = 169
AC = \(\sqrt{169}\) = 13
From the figuref ∆ABC,
Cos A = \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = \(\frac{12}{13}\), Cosec A = \(\frac{\mathrm{AC}}{\mathrm{BC}}\) = \(\frac{13}{5}\)
∴ Cos A = \(\frac{12}{13}\) and Cosec A = \(\frac{13}{5}\)

Question 5.
Evaluate the following.
i) Sin 60° + Cos 60°
Solution:
Sin 60° + Cos 60° = \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) = \(\frac{\sqrt{3}+1}{2}\)

ii) \(\frac{{Sin} 45^{\prime \prime}}{{Sin} 30^{\prime \prime}+{Cos} 60}\)
Solution:
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 3

iii) Tan2 30° + Cot2 45° – Cos 60
Solution:
Tan2 30° + Cot2 45° – Cos 60
= \(\left(\frac{1}{\sqrt{3}}\right)^2\) + (1)2 – \(\frac{1}{2}\)
= \(\frac{1}{3}\) + \(\frac{1}{1}\) – \(\frac{1}{2}\)
= \(\frac{2+6-3}{6}\)
= \(\frac{5}{6}\)

iv) 2 Tan2 45° + Sin2 60° – Cos2 30°
Sol:
2 Tan2 45° + Sin2 60° – Cos2 30°
= 2(1)2 + \(\left(\frac{\sqrt{3}}{2}\right)^2\) – \(\left(\frac{\sqrt{3}}{2}\right)^2\)
= 2 × 1 + \(\frac{3}{4}\) – \(\frac{3}{4}\) = 2

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

v) Cot2 30° + 4 Sin2 45° + 3 Cosec 60°
Solution:
Cot2 30° + 4 Sin2 45° + 3 Cosec 60°
= (\(\sqrt{3}\))2 + 4\(\left(\frac{1}{\sqrt{2}}\right)^2\) + 3\(\left(\frac{2}{\sqrt{3}}\right)^2\)
= 3 + 4 × \(\frac{1}{2}\) + 3 × \(\frac{4}{3}\)
= 3 + 2 + 4
= 9

vi) \(\sqrt{2}\) Sin 45° + Cos 90° + Sin 90°
Solution:
\(\sqrt{2}\) .Sin 45° + Cos 90° + Sin 90°
= \(\sqrt{2}\) . \(\frac{1}{\sqrt{2}}\) + 0 + 1
= 1 + 1
= 2

Question 6.
Evaluate Cos 60°, Cos 30° – Sin 60° Sin 30°
What is the value of Cos (60° + 30°) ? What can you conclude ?
Solution:
Take Cos 60° Cos 30° – Sin 60°. Sin 30°
= \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) – \(\frac{\sqrt{3}}{2}\) . \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) – \(\frac{\sqrt{3}}{4}\)
= 0 …………….. (1)
Now take Cos(60° + 30°) = Cos (90°) = 0 ………….. (2)
From equations (1) and (2), I conclude that
Cos(60° + 30°) = Cos 60° . Cos 30° – Sin 60°. Sin 30°
i.e.,Cos(A + B) = Cos A . Cos B – Sin A . Sin B.

Question 7.
Is it right to say Sin (60° – 30°) = Sin 60°. Cos 30° – Cos 60°. Sin 30° ?
Solution:
LHS = Sin(60° – 30°)
= Sin 30° = \(\frac{1}{2}\)
RHS = Sin 60°. Cos 30° – Cos 60°. Sin 30°
= \(\frac{\sqrt{3}}{2}\) . \(\frac{\sqrt{3}}{2}\) – \(\frac{1}{2}\) . \(\frac{1}{2}\)
= \(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{-1}{4}\)
= \(\frac{2}{4}\)
= \(\frac{1}{2}\)
Yes, it is right to say
Sin(60° – 30°) = Sin 60°. Cos 30° – Cos 60°. Sin 30°
i.e., Sin (A – B) = Sin A Cos B – Cos A. Sin B.

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 8.
Is it right to say that Cos (A + B) = Cos A + Cos B ?
Solution:
Take A = 60°, B = 30°
Then Cos(A + B) = Cos (60° + 30°)
= Cos 90° = 0
Cos A + Cos B = Cos 60° + Cos 30°
= \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\)
= \(\frac{1+\sqrt{3}}{2}\)
∴ Cos (A + B) ≠ Cos A + Cos B
It is not right to say that Cos (A + B) ≠ Cos A + Cos B

Question 9.
i) \(\frac{{Sin} 66^{\prime \prime}}{{Cos} 24 “}\)
ii) \(\frac{{Sin} 18^{\prime \prime}}{{Cos} 72^{\prime \prime}}\)
iii) \(\frac{{Tan} 80^{\prime \prime}}{{Cot} 10^{\prime \prime}}\)
iv) \(\frac{{Sec} 69^{\prime \prime}}{{Cosec} 21 “}\)
v) \(\frac{{Tan} 54 “}{{Cot} 36^{\prime \prime}}\)
Solution:
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 4
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 5

Question 10.
Find the value of
i) Sin 75° – Cos 15°
ii) Sec 23° – Cosec 67°
iii) Sec 70° – Sec 20°
iv) Tan 68° – Tan 22°
Solution:
i) Given Sin 75° – Cos 15°
= Sin 75° – Cos(90 – 75°)
= Sin 75° – Sin 75°
= 0

ii) Given Sec 23° – Cosec 67°
= Sec 23° – Cosec (90 – 23°)
= Sec 23° – Sec 23°
= 0

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

iii) Sec 70° Sec 20°
= Sin 70° . \(\frac{1}{\cos 20 “}\) = \(\frac{{Sin} 70^{\prime \prime}}{{Cos}\left(90^{\prime \prime}-70^{\prime \prime}\right)}\)
= \(\frac{{Sin} 70 “}{{Cos}(90 “-70 “)}\) = \(\frac{{Sin} 70^{\prime \prime}}{{Sin} 70^{\prime \prime}}\) = 1

iv) Given Tan 68°. Tan 22°
Tan 68° . Tan(90° – 68°)
Tan 68° . Cot 68°
= \(\frac{{Tan} 68^{\prime \prime}}{{Tan} 68^{\prime \prime}}\) = 1

Question 11.
If Cot 2A = Tan (A – 18), when 2A is an acute angle, find the value of A.
Solution:
Given that Cot 2A = Tan (A – 18°)
⇒ Tan (90° – 2A) = Tan (A – 18°) [∵ Cot θ = Tan (90 – θ)]
⇒ 90 – 2A = A – 18°
⇒ 90 + 18° = A + 2A
⇒ 3A = 108°
⇒ A = \(\frac{108 “}{3}\) = 36°
Hence, the value of A is 36°.

Question 12.
If Cos 4A = Sin(A – 20), when 4A is an acute angle, find the value of A.
Solution:
Given that Cos 4A = Sin(A – 20°)
⇒ Sin(90° – 4A) = Sin(A – 20°)
[∴ Cos 0 = Sin(90 – 0)]
⇒ 90 – 4A = A – 20°
⇒ 90 + 20 = A + 4A
⇒ 110° = 5A
⇒ A = \(\frac{110 “}{15}\) = 22°
Hence, the value of A is 22°.

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 13.
If Sin θ + Cosec θ = 2, find the value of Sin2 θ + Cosec2 θ
Solution:
Given Sin θ + Cosec θ = 2
Squaring on both sides
(Sin θ + Cosec)2 = 22
⇒ Sin2 θ + Cosec2 θ + 2 . Sin θ . Cosec θ = 4
⇒ Sin2 θ + Cosec2 θ + 2 . Sin θ . \(\frac{1}{{Sin} \theta}\) = 4
⇒ Sin2 θ + Cosec2 θ + 2 = 4
⇒ Sin2 θ + Cosec2 θ + 4 – 2
⇒ Sin2 θ + Cosec2 θ = 2

Question 14.
Show that \(\frac{{Tan} A+{Cot} B}{{Tan} B+{Cot} A}\) = Tan A . Cot B.
Solution:
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 6
= Cot B . Tan A
= Tan A . Cot B = RHS
LHS = RHS

Question 15.
Show that (Sin θ + Cos θ)2 – (Sin θ – Cos θ)2 = 4 Sin θ Cos θ
Solution:
LHS = (Sin θ + Cos θ)2 – (Sin θ – Cos θ)2
= (Sin2 θ + Cos2 θ + 2 Sin θ . Cos θ) – (Sin2 θ + Cos2 θ – 2 Sin θ . Cos θ)
= (2 Sin θ + Cos θ) – (-2 Sin θ . Cos θ)
= 2 Sin θ + Cos θ + 2 Sin θ . Cos θ
= 4 Sin θ . Cos θ
= RHS
∴ LHS = RHS

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 16.
Find the value of \(\frac{{Tan}^2 60^{\prime \prime}+4 {Sin}^2 45^{\prime \prime}+3 {Sec}^2 30^{\prime \prime}+10 {Cos}^2 90^{\prime \prime}}{{Cosec}^2 60^{\prime \prime}+{Cos} 60^{\prime \prime}-{Cot}^2 60^{\prime \prime}}\)
Solution:
Put the following values in the given problem.
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 7

Question 17.
Show that \(\frac{1}{{Sin} \theta}\) – Sin θ = Cot θ. Cos θ
Solution:
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 8

Question 18.
Show that \(\frac{{Tan}^2 q}{1+{Sec} q}=\frac{1-{Cos} q}{{Cos} q}\)
Solution:
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 9
∴ LHS = RHS

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 19.
Evaluate : log4 (1 + tan2 45°)2
Solution:
∴ log4 (1 + 1)2
= log4 = 1

Question 20.
Is it true to say that cos (60°+30°) = cos 60° cos 30° + sin 60° sin 30°
Solution:
Here L.H.S = cos (60° + 30°) = cos 90° = 0
R.H.S – cos 60° cos 30° + sin 60° sin 30°
= \(\frac{1}{2}\) × \(\frac{\sqrt{3}}{2}\) + \(\frac{\sqrt{3}}{2}\) × \(\frac{1}{2}\)
= \(\frac{\sqrt{3}}{4}\) + \(\frac{\sqrt{3}}{4}\)
= \(\frac{\sqrt{3}}{4}\)
= \(\frac{\sqrt{3}}{2}\)
Here L.H.S ≠ R.H.S, so it’s not justify.

Question 21.
In ∆ABC, ∠C = 90° If BC + CA = 17 cm; BC – CA = 7 cm. Find
(i) sin A
(ii) sin B
Solution:
Here, given ∠C = 90°; BC + CA = 7 cm;
BC – CA = 7 cm
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 10
BC = \(\frac{24}{2}\) = 12 cm; figure script pno.4
BC = 12 cm
We apply BC = 12 cm. In BC + CA = 17 cm
Then 12 + CA = 17
CA = 17 – 12 = 5 cm
We know from ∆ABC, AB2 = BC2 + CA2
AB = \(\sqrt{\mathrm{BC}^2+\mathrm{CA}^2}\)
AB = \(\sqrt{12^2+5^2}\) = \(\sqrt{144+5}\)
AB = \(\sqrt{169}\) = 13
We know = sin A = \(\frac{\mathrm{BC}}{\mathrm{AB}}\) = \(\frac{12}{13}\)
sin B = \(\frac{\mathrm{AC}}{\mathrm{AB}}\) = \(\frac{5}{13}\)

TS 10th Class Maths Important Questions Chapter 11 Trigonometry

Question 22.
Find the value of cos2 1° + cos2 2° + cos2 3° + ……………. + cos2 90°
Solution:
We know cos2 (90° – 89°) + cos2 (90° – 88°) + …………… + cos2 89° + cos2 90°
Here total 90° terms is there we know sin2θ + cos2θ = 1
= (sin2 89° + cos2 89) + (sin2 89 + cos2 88) + ………….. 44 terms.
= 44(1) + \(\frac{1}{\sqrt{2}}\) + 1
= 45 + \(\frac{1}{\sqrt{2}}\)

Question 23.
If cosec θ + cot θ = k then prove that cos θ = \(\frac{\mathrm{k}^2-1}{\mathrm{k}^2+1}\).
Solution:
Given cosec θ + cot θ = k ………….. (1)
We know cosec2θ – cot2θ = 1
∴ (cosec θ + cot θ) (cosec θ – cot θ) = 1 k(cosec θ – cot θ) = 1
k(cosec θ – cot θ) = 1
cosec θ – cot θ = …………….. (2)
TS 10th Class Maths Important Questions Chapter 11 Trigonometry 11

Leave a Comment