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## 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.

Now, by using the pythagoras theorem, we have

AC^{2} = AB^{2} + BC^{2}

= (4k)^{2} + (3k)^{2} = 25 k^{2}

AC = 5k

So,

Now, we can write all the trigonometric ratios using their definitions.

Question 2.

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.

Which is the RHS of the identity, we are required to prove.

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)

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.

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:

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

cos^{2}A + sin^{2}A = 1, therefore,

cos^{2}A = 1 – sin^{2}A

i.e., cos A = \(\pm \sqrt{1-\sin ^{2} A}\)

This gives cos A = \(\sqrt{1-\sin ^{2} A}\)

Question 7.

Evaluate \(\frac{\tan 65^{\circ}}{\cot 25^{\circ}}\)

Solution:

We know:

cot A = tan(90° – A)

So,

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°

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.

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)