Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

Question 1.
Given tan A = \(\frac{4}{3}\), find the other trigonometric ratios of the angle A.
Solution:
Let us first draw a right ∆ABC.
Now, we know that tan A = \(\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}\)
Therefore, if BC = 4k, then AB = 3k, where k is a positive number.
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 1
Now, by using the pythagoras theorem, we have
AC2 = AB2 + BC2
= (4k)2 + (3k)2 = 25 k2
AC = 5k
So,
Now, we can write all the trigonometric ratios using their definitions.
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 2

Question 2.
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 3
Solution:
Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by cos θ.
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 4
Which is the RHS of the identity, we are required to prove.

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

Question 3.
Prove that sec A (1 – sin A) (sec A + tan A) = 1.
Solution:
LHS = sec A(1 – sin A)(sec A + tan A)
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 5

Question 4.
In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
Solution:
In ABC, tan A = \(\frac{B C}{A B}\)
Let AB = BC = k, where k is a positive number.
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 50

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

Question 5.
If sin (A – B) = \(\frac{1}{2}\), cos (A + B) = \(\frac{1}{2}\), 0° < A + B ≤ 90°, A > B, find A and BC
Solution:
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 6
We get,
A = 45° and B = 15°

Question 6.
Express the ratios cos A, tan A and sec A in terms of sin A.
Solution:
Since
cos2A + sin2A = 1, therefore,
cos2A = 1 – sin2A
i.e., cos A = \(\pm \sqrt{1-\sin ^{2} A}\)
This gives cos A = \(\sqrt{1-\sin ^{2} A}\)
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 7

Question 7.
Evaluate \(\frac{\tan 65^{\circ}}{\cot 25^{\circ}}\)
Solution:
We know:
cot A = tan(90° – A)
So,
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 9

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

Question 8.
Since sin 3A = cos(A – 26°), where 3A is an acute angle, find the value at A.
Solution:
We are given that sin 3A = cos (A – 26°) ….. (1)
Since sin 3A = cos(90° – 3A) we can write (1) as cos(90° – 3A) = cos(A – 26°)
Since 90° – 3A and A – 26° are both acute angles.
90° – 3A = A – 26°
which gives A = 29°

Question 9.
Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
cot 85° +cos 75°
= cot(90° – 5°) + cos(90° – 15°)
= tan 5° + sin 15°

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions

Question 10.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides at the river are 30° and 45°, respectively. If the bridge is at a height at 3 m from the banks, find the width at the river.
Solution:
A and B represent points on the bank on opposite sides at the river, so that AB is the width of the river. P is a point on the bridge at a height of 3m i.e., DP = 3 m. We are interested to determine the width at the river which is the length at the side AB of the ∆APB.
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 10
Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Additional Questions 11
Also, in right ∆PBD,
B = 45°
So, BD = PD = 3 m
Now, AB = BD + AD
= 3 + \(3 \sqrt{3}\) = 3(1 + \(\sqrt{3}\))m
Therefore, the width at the river is 3(\(\sqrt{3}\) + 1)