Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers
Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.
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Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type
Question 1.
From the given figure, find the value of x:
Answer:
In the given fig., only one side is known which is hypotenuse and side to be evaluated is BC which is perpendicular with reference to given angle ZA = 30°.
∴ sin 30° = \(\frac{x}{15}\) ⇒ x = 15 sin 30°
x = 15 × \(\frac{1}{2}\) = 7\(\frac{1}{2}\) cm
Question 2.
If tan A = cot B, prove that A + B = 90°.
Answer:
∵ tan A = cot B
∴ tan A = tan (90°-B)
⇒ A = 90°-B
⇒ A + B = 90°. [Hence proved]
Question 3.
Evaluate cos 60° sin 30° + sin 60° cos 30°. (AI CBSE 2010)
Answer:
We have: cos 60° . sin 30° + sin 60°. cos 30°
Question 4.
If cos A = \(\frac{2}{5}\), find the value of 4 + 4 tan2A [CBSE Sample Paper-2017]
Answer:
sec A = \(\frac{1}{\cos \mathrm{A}}=\frac{5}{2}\)
4 + 4 tan2 A = 4 (1 + tan2 A)
= 4 (sec2 A) = 4 \(\left(\frac{5}{2}\right)^2\) = 25
Question 5.
If 3x = sec θ and 9\(\left(x^2-\frac{1}{x^2}\right)\) = tan θ, then find \(\).
Answer:
Question 6.
What is the value of (cos2 67° – sin2 23°)?
Answer:
∵ 67° + 23° = 90°
∴ cos2 67° – sin2 23° = cos2 (90° – 23°) – sin223°
= sin2 23° – sin2 23°
= 0
Question 7.
Find the value of sin 38° – cos 52°? [CBSE 2016]
Answer:
sin 38° – cos 52°
= sin38° – cos (90° – 38°) [∵ cos (90° – θ = sin θ]
= sin 38° – sin 38°
= 0
Question 8.
If 7 sin2 θ + 3 cos2 θ = 4, then find the value of tan θ.
Answer:
7 sin2 θ + 3 cos2 θ = 4
Dividing both sides by cos2 θ
\(\frac{7 \sin ^2 \theta}{\cos ^2 \theta}\) + 3 = \(\frac{4}{\cos ^2 \theta}\)
7 tan2 θ + 3 = 4 sec2 θ
= 4(1 + tan2 θ)
⇒ 3 tan2 θ = 1
tan θ = \(\frac{1}{\sqrt{3}}\)
Question 9.
Write the value of \(\frac{5}{\cot ^2 \theta}-\frac{5}{\cos ^2 \theta}\)
Answer:
\(\frac{5}{\cot ^2 \theta}-\frac{5}{\cos ^2 \theta}\) = 5 tan2 θ – 5 sec2 θ
= 5(tan2 θ – sec2 θ)
= – 5(sec2 θ – tan2 θ)
= – 5(1)
= – 5
Question 10.
Given that tan θ = \(\frac{1}{\sqrt{3}}\) then find the value of \(\frac{{cosec}^2 \theta-\cot ^2 \theta}{{cosec}^2 \theta+\sec ^2 \theta}\).
Answer:
Question 11.
If tan A = cot (A + 10), find A. [CBSE Delhi 2016]
Answer:
tan A = cot (A + 10)
⇒ cot (90° – A) = cot (A + 10) [ ∵ cot (90 – θ) = tan θ]
⇒ 90° – A = A + 10
⇒ 90° – 10 = A + A
⇒ 80° = 2A
\(\frac{80^{\circ}}{2}\) = A
∴ A = 40°
Question 12.
Express sin 67° + cos 75° in terms of t-ratios of angles between 0° and 45°.
Answer:
∵ 67 = 90 – 23 and 75 = 90 – 15
∴ sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°.
Question 13.
Find the value of sin (60° + θ) – cos (30° – θ).
Answer:
Observe (60° + θ) = 90° – (30° – θ)
∴ sin (60°+ θ) – cos (30° – θ)
= sin {90° – (30° – θ)} – cos (30° – θ)
= cos (30° – θ) – cos (30° – θ)
= 0
Introduction to Trigonometry Class 10 Extra Questions Short Answer Type-1
Question 1.
Express cot 85° + cos 75° in terms of trigonometric ratio of angles between 0° and 45°.
Answer:
We have
cot 85° + cos 75°
= cot (90° – 5°) + cos (90° – 15°)
= tan 5° + sin 15°
[∵ cot (90° – 0) = tan θ
and cos (90° – θ) = sin θ]
which is the required result.
Question 2.
Write the simplest value of \(\frac{\cos ^2 \theta}{\sin \theta}\) + sin θ
Answer:
\(\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta}\) = \(\frac{1}{\sin \theta}\) = cosec θ
Question 3.
If sin 3θ = cos (θ – 6)° and 30 and (θ – 6)° are acute angles, find the value of θ.
Answer:
We have:
sin 30 = cos (θ – 6)°
= sin [90°- (θ – 6)°]
[∵ sin (90° – θ) = cos θ]
⇒ 3θ = 90° – (θ – 6)°
⇒ 3θ = 90 – θ + 6
⇒ 3θ + θ = 96
⇒ 4θ = 96
⇒ θ = \(\frac{96}{4}\) = 24
Thus θ = 24°.
Question 4.
Prove the following identity:
\(\left(1+\frac{1}{\tan ^2 A}\right)\left(1+\frac{1}{\cot ^2 A}\right)\) = \(\frac{1}{\cos ^2 A-\cos ^4 A}\) [CBSE 2016]
Answer:
As, 1 + cot2 A = cosec2 A
1 + tan2 A = sec2 A
sin2 A + cos2 A = 1
L.H.S = R.H.S.
Hence Proved.
Question 5.
If tan θ = cot (30° + θ), find the value of θ.
Answer:
We have:
tan θ = cot (30° + θ)
= tan [90° – (30° + θ)]
= tan [90° – 30°- θ]
= tan (60° – θ)
⇒ 0 = 60° – θ
⇒ θ + θ = 60°
⇒ 20 = 60°
⇒ θ = \(\frac{60^{\circ}}{2}\) = 30°
∴ θ = 30°.
Question 6.
If 3 cot A = 4, find the value of \(\frac{{cosec}^2 A+1}{{cosec}^2 A-1}\)
Answer:
Question 7.
Without using trigonometric tables, prove that \(\frac{\cos ^2 49^{\circ}+\cos ^2\left(90^{\circ}-49^{\circ}\right)}{\sin ^2 31^{\circ}+\sin ^2\left(90^{\circ}-31^{\circ}\right)}\) + 2 tan 35° tan 55° = 3.
Answer:
Question 8.
Prove that: \(\sqrt{\frac{\sec \theta-1}{\sec \theta+1}}+\sqrt{\frac{\sec \theta+1}{\sec \theta-1}}\) = 2 cosec θ
Answer:
Question 9.
Show that: tan 10° tan 15° tan 75° tan 80° = 1
Answer:
We have:
LHS = tan 10° tan 15° tan 75° tan 80°
= tan (90° – 80°) tan 15° tan (90° -15°) tan 80°
= cot 80° tan 15° cot 15° tan 80°
= (cot 80° × tan 80°) × (tan 15° × cot 15°)
= 1 × 1 = 1 = RHS
Question 10.
Evaluate cosec (65° + θ) – sec (25° – θ) – tan (55° – θ) + cot (35° + θ). [C.B.S.E. 2000]
Answer:
Here observe that:
(65° + θ) + (25° – θ) = 90°
⇒ 65° + θ = 90° – (25° – θ)
(55° – θ) + (35° + θ) = 90°
⇒ 35° + θ = 90° – (55° – θ)
∴ Given expression
cosec [9θ° – (25 – θ)] – sec (25° – θ) – tan (55° – θ) + cot [9θ° – (55° – θ)]
= sec (25° – θ) – sec (25° – θ) – tan (55° – θ) + tan (55° – θ)
[∵ cosec (9θ° – θ) = sec θ cot (9θ° – θ) = tan θ
= 0 – 0
= θ
Question 11.
Prove the following identities:
(i) (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ
(ii) (sin θ + sec θ)2 + (cos θ + cosec θ)2 = (1 + sec θ cosec θ)2 [CBSE 2001C]
Answer:
(i) LHS = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2 θ + cosec2 θ + 2 sin θ cosec θ) + (cos2 θ + sec2 θ + 2 cos θ sec θ)
= (sin2 θ + cos2 θ) + (cosec2 θ + sec2 θ) + 2(sin θ cosec θ + cos θ sec θ)
= 1 + (1 + cot2 θ + 1 + tan2 θ) + 2(1 + 1)
= 3 + cot2 θ + tan2 θ + 4
[∵ sin θ cosec θ = 1 = cos θ sec θ]
= (3 + 4) + tan2 θ + cot2 θ
= 7 + tan2 θ + cot2 θ
= RHS
Hence Proved.
(ii) LHS = (sin θ + sec θ)2 + (cos θ + cosec θ)2
= (sin2 θ + sec2 θ + 2 sin θ sec θ) + (cos2 θ + cosec2 θ + 2 cos θ cosec θ)
= (sin2 θ + cos2 θ) + (sec2 θ + cosec2 θ) + 2 sin θ sec θ + 2 cos θ cosec θ
= 1 + (1 + tan2 θ + 1 + cot2 θ) + \(\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)\)
= 1 + (tan2 θ + cot2 θ + 2 tan θ cot θ) + 2\(\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\right)\)
= 1 + (tan θ + cot θ)2 + 2 cosec θ sec θ
= 1 + \(\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)^2\) + 2 cosec θ sec θ
= 1 + sec2 θ cosec2 θ + 2 cosec θ sec θ
= (1 + sec θ cosec θ)2
= RHS
Hence proved
Question. 12.
If cos θ + sin θ = √2 cos θ, show that cos θ – sin θ = √2 sin θ.
Answer:
Given, cos θ + sin θ = √2 cos θ ……. (i)
Squaring both sides
cos2 θ + sin2 θ + 2 sin θ cos θ = 2 cos2 θ
⇒ cos2 θ – sin2 θ = 2 sin θ cos θ
⇒ (cos θ – sin θ) (cos θ + sin θ) = 2 sin θ cos θ
⇒ (cos θ – sin θ) (√2 cos θ) = 2 sin θ cos θ (Using (i))
⇒ cos θ – sin θ = \(\frac{2 \sin \theta \cos \theta}{\sqrt{2} \cos \theta}\)
⇒ cos θ – sin θ = √2 sin θ
Introduction to Trigonometry Class 10 Extra Questions Short Answer Type-2
Question 1
If 7 sin2 A + 3 cos2 A = 4, show that tan A = \(\frac{1}{\sqrt{3}}\). [CBSE Delhi 2016]
Answer:
7 sin2 A + 3 cos2 A = 4
Dividing both sides by cos2 A
⇒ 7 tan2 A + 3 = 4 sec2 A
⇒ 7 tan2 A + 3 = 4(1 + tan2A) [∵sec2θ = 1 + tan2 θ]
⇒ 7 tan2 A + 3 – 4(1 + tan2 A) = 0
⇒ 7 tan2 A + 3 – 4 – 4 tan2 A = 0
⇒ 3 tan2 A – 1 = 0
⇒ 3 tan2 A = 1
⇒ tan2 A = \(\frac{1}{3}\)
∴ tan A = \(\frac{1}{\sqrt{3}}\)
∴ Hence proved.
Question 2.
Prove (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A. [CBSE Delhi 2016]
Answer:
We take,
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= (sin2 A + cosec2 A + 2 sin A cosec A) + (cos2 A + sec2 A + 2 cos A sec A)
= sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
= (sin2 A + cos2 A) + (cosec2 A + sec2 A) + 2 (sin A cosec A + cos A sec A)
= 1 + (1 + cot2 A + 1 + tan2 A) + 2(1 + 1)
= 3 + cot2 A + tan2 A + 4
[∵ sin θ cosec θ = 1
cos θ sec θ = 1]
= (3 + 4) + (cot2 A + tan2 A)
= 7 + tan2 A + cot2 A
∴ = R.H.S.
∴ Hence proved.
Question 3.
If sin θ + cos θ = √2, then evaluate tan θ + cot θ [CBSE Sample Paper-2017]
Answer:
sin θ + sos θ = √2
⇒ (sin θ + cos θ)2 = (√2)2
⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 2
⇒ 1 + 2 sin θ cos θ = 2
⇒ sin θ cos θ = \(\frac{1}{2}\) ………… (i)
we know, sin2 θ + cos2 θ = 1 ……………… (ii)
Dividing (ii) by (i) wet get
⇒ tan θ + cot θ = 2
Question 4.
If sec 3A = cosec (A – 10°) where 3A is an acute angle, find the value of A.
Answer:
sec 3A = cosec (A -10°)
sec 3A = sec {90° – (A – 10)}
[∵sec (90° – θ) = cosec θ]
⇒ sec 3A = sec (100° – A)
⇒ 3A = 100° – A
⇒ 3A + A = 100°
4A = 100°
A = \(\frac{100^{\circ}}{4}\) = 25°
Question 5.
Prove that:
sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A [C.B.S.E. 2000 (F)]
Answer:
LHS = sin A (1 + tan A) + cos A (1 + cot A)
Question 6.
Evaluate
\(\frac{{cosec}^2 63^{\circ}+\tan ^2 24^{\circ}}{\cot ^2 66^{\circ}+\sec ^2 27^{\circ}}\) + \(\frac{\sin ^2 63^{\circ}+\cos 63^{\circ} \sin 27^{\circ}+\sin 27^{\circ} \sec 63^{\circ}}{2\left({cosec}^2 65^{\circ}-\tan ^2 25^{\circ}\right)}\) [CBSE Sample Paper-2017]
Answer:
Question 7.
If cos2 θ – sin2 θ = tan2 Φ, prove that
cos Φ = \(\frac{1}{\sqrt{2} \cos \theta}\)
Answer:
Consider cos2 θ – sin2 θ = tan2 Φ
⇒ cos2 θ – (1 – cos2 θ) = sec2 Φ – 1
[∵ sin2 θ + cos2 θ = 1, sec2 Φ = 1 + tan2 Φ]
⇒ 2cos2 θ – 1 = sec2Φ – 1
⇒ 2 cos2 θ = sec2 Φ
Taking square root, we get
√2 cos θ = sec Φ
⇒ √2 cos θ = \(\frac{1}{\cos \phi}\)
⇒ cos Φ = \(\frac{1}{\sqrt{2} \cos \theta}\)
Question 8.
If tan 2A = cot (A -18°), where 2A is an acute angle, find the value of A. [CBSE 2018]
Answer:
We know that tan θ = cot (90 – θ)
∴ tan 2A = cot (A – 18°)
⇒ cot (90 – 2A) = cot (A – 18°)
⇒ 90° – 2A = A – 18°
⇒ 3A = 108°
A = 36°
Question 9.
Without using trigonometric tables, evaluate the following:
\(\frac{\sec 39^{\circ}}{{cosec} 51^{\circ}}+\frac{2}{\sqrt{3}}\) = tan 17° tan 38° tan 60° tan 52° tan 73° – 3 (sin2 31° + sin2 59°) [CBSE 2006(C)]
Answer:
= 1 + 2 (tan 17° cot 17°) (tan 38° cot 38°) – 3 × 1
= 1 + 2 × 1 × 1 – 3 = 3 – 3 = 0 [∵ tan θ cot θ = 1]
Question 10.
Prove that: a2 + b2 = x2 + y2 when a cos θ – b sin θ = x and a sin θ + b cos θ = y.
Answer:
RHS = x2 + y2
= [a cos θ – b sin θ]2 + [a sin θ + b cos θ]2
= a2 cos2 θ + b2 sin2 θ – 2ab sin θ cos θ + a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ
= a2 cos2 θ + b2 sin2 θ + a2 sin2 θ + b2 cos2 θ
= a2 [cos2 θ + sin2 θ] + b2 [sin2 θ + cos2 θ]
= a2[1] + b2[1] [ ∵ sin2 θ + cos2 θ = 1]
= a2 + b2
= LHS
Hence proved
Introduction to Trigonometry Class 10 Extra Questions Long Answer Type 1
Question 1.
If 3 tan A = 4, check whether = \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\) = cos2A – sin2A or not? [CBSE Delhi 2016]
Answer:
Now, take R.H.S. = cos2 A – sin2 A
Firstly, we find the value of cos A and sin A.
By using Pythagoras Theorem.
(AC)2 = (AB)2 + (BC)2
⇒ (AC)2 = (3)2 + (4)2 = 9 + 16 = 25
AC = √25
∴ AC = 5
∴ cos2 A – sin2 A = \(\frac{-7}{25}\)
∴ Hence, L.H.S. = R.H.S.
Hence proved.
Question 2.
If x = a sin θ and y = b tan θ, then prove that \(\frac{a^2}{x^2}-\frac{b^2}{y^2}\) = 1.
Answer:
Question 3.
Prove that \(\frac{2}{\cos ^2 \theta}-\frac{1}{\cos ^4 \theta}-\frac{2}{\sin ^2 \theta}+\frac{1}{\sin ^4 \theta}\) = cot4 θ – tan4 θ
Answer:
= 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ
= 2(1 + tan2 θ) – (1 + tan2 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)2
= (1 + tan2 θ) [2 – (1 + tan2 θ)] – (1 + cot2 θ) [2 – (1 + cot2 θ)]
= (1 + tan2 θ) (1 – tan2 θ) – (1 + cot2 θ) (1 – cot2 θ)
= 1 – tan4 θ – (1 – cot4 θ) = cot4 θ – tan4 θ
= RHS
Hence proved
Question 4.
Evaluate sec 41° sin 49° + cos 49° cosec 41° + \(\frac{2}{\sqrt{3}}\) = tan 20° tan 60° tan 70° – 3 (cos2 45° – sin2 90°).
Answer:
The given expression is
sec 41° sin 49° + cos 49° cosec 41° + \(\frac{2}{\sqrt{3}}\) tan 20° tan 60° tan 70° – 3 (cos2 45° – sin2 90°)
= sec 41° × sin (90° – 41°) + cos (90° – 41°) × cosec 41° + \(\frac{2}{\sqrt{3}}\) tan 20° × tan 70° × tan 60° – 3 (cos245° – (1)2)
= sec 41° × cos 41° + sin 41° × cosec 41°
Question 5.
Determine the value of x such that
2 cosec2 30° + x sin2 60° – \(\frac{3}{4}\) tan2 30° = 10.
Answer:
We have,
2 cosec22 30° + x sin2 60° – \(\frac{3}{4}\) tan2 30° = 10
Introduction to Trigonometry Class 10 Extra Questions HOTS
Question 1.
Show that cosec2 θ + sec2 θ can never be less than 2.
Answer:
Let, if possible, cosec2 θ + sec2 θ < 2
⇒ (1 + cot2 θ) + (1 + tan2 θ) < 2
⇒ cot2 θ + tan2 θ + 2 < 2
⇒ cot2 θ + tan2 θ < θ
But sum of squares of two real numbers is always non negative.
∴ cot2 θ + tan2 θ < 0 is not possible.
Therefore, our supposition is wrong.
Hence cosec2 θ + sec2 θ can never be less than 2.
Question 2.
If sin θ + sin2θ + sin3θ = 1, then find the value of cos6 θ – 4 cos4 θ + 8 cos2 θ.
Answer:
We have, sin θ + sin3 θ = 1- sin2 θ
⇒ sin θ (1 + sin2 θ) = cos2 θ
⇒ sin θ (2 – cos2 θ) = cos2 θ
Squaring both sides, we get
sin2 θ (2 – cos2 θ)2 = cos4 θ
⇒ (1 – cos2 θ) (4 + cos4 θ – 4 cos2 θ) = cos4 θ
⇒ 4 + cos4 θ – 4 cos2 θ – 4 cos2 θ – cos6 θ + 4 cos4 θ = cos4 θ
⇒ cos6 θ – 4 cos4 θ + 8 cos2 θ = 4.
Question 3.
If 7 sin2 θ + 3 cos2 0 = 4, then, show that tan θ = \(\frac{1}{\sqrt{3}}\), where θ is an acute angle.
Answer:
7 sin2 θ + 3 cos2 θ = 4
⇒ 7 sin2 θ + 3 cos2 θ = 4 × 1 = 4 × (sin2 θ + cos2 θ)
⇒ 3 sin2 θ = cos2 θ
Question 4.
Prove that: 3 (sin θ – cos θ)4 + 6 (sin θ + cos θ)2 + 4 (sin6 θ + cos6 θ) = 13.
Answer:
(sin θ – cos θ)4
= [(sin θ – cos θ)2]2
= [sin2θ + cos2θ – 2 sin θ cos θ]2
= [1 – 2 sin θ cos θ]2
= 1 + 4 sin2θ cos2θ – 4 sin θ cos θ Also (sin θ + cos θ)2
= sin2 θ + cos2 θ + 2 sin θ cos θ and (sin6 θ + cos6 θ)
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ) [(sin2θ)2 – sin2 θ cos2 θ + (cos2 θ)2]
= (1) [sin4 θ – sin2 θ cos2 θ + cos4 θ]
= [(sin2θ + cos2θ)2– 2 sin2 θ cos2 θ – sin2 θ cos2 θ]
= 1 – 3 sin2 θ cos2 θ
∴ LHS
= 3 (1 + 4 sin2 θ cos2 θ – 4 sin θ cos θ) + 6 (1 + 2 sin θ cos θ) + 4 (1 – 3 sin2 θ cos2 θ)
= 3 + 6 + 4 = 13
= RHS Hence proved
Multiple Choice Questions
Choose the correct option out of four given in each of the following:
Question 1.
The value of 0, for sin2 0 = 1, where 0° < θ < 90° is
(a) 30°
(b) 60°
(c) 45°
(d) 135°
Answer:
(c) 45°
Question 2.
If cos A = \(\frac{4}{5}\), then value of tan A is
(a) \(\frac{3}{5}\)
(b) \(\frac{3}{4}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{5}{3}\)
Answer:
(b) \(\frac{3}{4}\)
Question 3.
In ∆ABC, ∠B = 90°. If AB = 14 cm and AC = 50 cm then tan A equals:
(a) \(\frac{24}{25}\)
(b) \(\frac{24}{7}\)
(c) \(\frac{7}{24}\)
(d) \(\frac{25}{24}\)
Answer:
(c) \(\frac{7}{24}\)
Question 4.
If 4 cot θ = 3 then \(\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}\) is equal to :
(a) \(\frac{5}{6}\)
(b) \(\frac{1}{7}\)
(c) \(\frac{6}{7}\)
(d) \(\frac{3}{4}\)
Answer:
(b) \(\frac{1}{7}\)
Question 5.
Value of cosec2 27° – tan2 63° is equal to :
(a) 1
(b) 0
(c) – 1
(d) 90
Answer:
(c) – 1
Question 6.
The value of the expression [cosec (85° + θ) – sec (5° – θ) – tan (42° + θ) + cot (48° – θ)] is
(a) – 1
(b) 0
(c) 1
(d) \(\frac{3}{2}\)
Answer:
(b) 0
Question 7.
If A + B = 90° and cot B = \(\frac{3}{4}\) then tan A is equal to
(a) \(\frac{3}{4}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{1}{4}\)
(d) \(\frac{1}{3}\)
Answer:
(a) \(\frac{3}{4}\)
Question 8.
If ∆ABC is right angled at C, then value of cos(A + B) is
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) \(\frac{\sqrt{3}}{2}\)
Answer:
(a) 0
Question 9.
Maximum of \(\frac{1}{{cosec} \theta}\), 0° < θ < 90° is
(a) – 1
(b) 2
(c) 1
(d) \(\frac{1}{2}\)
Answer:
(c) 1
Question 10.
The value of tan 1° tan 2° tan 3° ………….. tan 89° is
(a) 0
(b) 1
(c) 2
(d) 0
Answer:
(b) 1
Question 11.
Given that sin α = \(\frac{\sqrt{3}}{2}\) and cos β = \(\frac{1}{2}\) then the value of (α + β) is
(a) 90°
(b) 30°
(c) 60°
(d) 120°
Answer:
(d) 120°
Question 12.
(1 – cos θ) (cot θ + cosec θ) is equal to :
(a) sin θ
(b) sec θ
(c) tan θ
(d) cot θ.
Answer:
(a) sin θ
Question 13.
If 3 cos θ = 1, then the value of \(\frac{6 \sin ^2 \theta+\tan ^2 \theta}{4 \cos \theta}\)
(a) 5
(b) 10
(c) -10
(d) 0
Answer:
(b) 10
Question 14.
If sec2 θ (1 + sin θ) (1 – sin θ) = k then value of k is
(a) 2
(b) 3
(c) θ
(d) 1
Answer:
(d) 1
Question 15.
If sin θ = \(\frac{1}{3}\), then value of (2 cot2 θ + 2) is
(a) 9
(b) 18
(c) 27
(d) 20
Answer:
(b) 18
Fill in the Blanks:
Question 1.
5 sin θ° + cos 9θ° = ___________ .
Answer:
0
Question 2.
cosec3 (9θ° – θ) cos3 (9θ° – θ) = ___________ .
Answer:
tan3θ
Question 3.
Value of tan 90° is ___________.
Answer:
undefined/∞
Question 4.
If sin θ = 1, then θ = ___________
Answer:
90°
Question 5.
1 + tan2 θ = sec2 θ is not valid for θ = ___________ .
Answer:
90°
Question 6.
2 tan2 45° + 2 cos2 45° – 3 sin 9θ° = ___________ .
Answer:
0
Question 7.
sin θ ___________ when θ increases from θ° to 9θ°.
Answer:
increases
Question 8.
cos θ ___________ when θ decreases from θ° to 9θ°.
Answer:
decreases
Question 9.
(1 + tan2 θ) (1 + sin θ) (1 – sin θ) = ___________ .
Answer:
1
Question 10.
\(\frac{\cos 58^{\circ}}{\sin 32^{\circ}}-\frac{\tan 31^{\circ}}{\cot 59^{\circ}}\) ___________ .
Answer:
0