Trigonometric Ratios NTSE JEE Foundation Important Solved Examples and Problems

ANAND CLASSES Study Material and Notes to Strengthen your Class 10 Mathematics with essential solved examples on Trigonometric Ratios for NTSE JEE Foundation. Learn step-by-step problem-solving techniques to score higher in exams.

Express the trigonometric ratios cos A, tan A and sec A in terms of sin A.

Solution:

We know that,

sin² A + cos² A = 1

So,

cos A = √(1 – sin² A)

tan A = sin A/cos A = sin A/ √(1 – sin² A)

sec A = 1/cos A = 1/ (√1 – sin² A)

Follow the Figure for the following problems

If A is an acute angle and sin A = 3/5, find all other trigonometric ratios of angle A (using trigonometric identities).

Solution:

Given, sin A = 3/5 and A is an acute angle

So, in ∆ABC we have ∠B = 90^o

And, AC = 5 and BC = 3

By Pythagoras theorem,

AB = √(AC² – BC²)

AB = √(5² – 3²) = √(25 – 9) = √16 = 4

Now,

cos A = AB/AC = 4/5

tan A = BC/AB = 3/4

cot A = 1/tan θ = 4/3

sec A = 1/cos θ = 5/4

cosec A = 1/sin θ = 5/3

If A is an acute angle and sec A = 17/8, find all other trigonometric ratios of angle A (using trigonometric identities).

Solution:

Given, sec A = 17/8 and A is an acute angle

So, in ∆ ABC we have ∠B = 90^o

And, AC = 17 and AB = 8

By Pythagoras theorem,

BC = √(AC² – AB²)

BC = √(17² – 8²) = √(289 – 64) = √225 = 15

Now,

sin A = BC/AC = 15/17

cos A = 1/sec A = 8/17

tan A = BC/AB = 15/8

cot A = 1/tan A= 8/15

cosec A = 1/sin A = 17/15

If tan A = 1/√3, find all other trigonometric ratios of angle A.

Solution:

Given, tan A = 1/√3

In the right ∆ ABC, we have ∠B = 90^o

tan A = BC/AB = 1/√3

So,

BC = 1 and AB = √3

By Pythagoras theorem,

AC = √(AB² + BC²) = √[(√3)² + (1)²]

AC = √(3 + 1) = √4 = 2

Hence,

sin A = BC/AC = ½

cos A = AB/AC = √3/2

cot A = 1/tan A = √3

sec A = 1/cos A = 2/√3

cosec A = 1/sin A = 2/1 = 2

If 12 cosec θ = 13, find the value of (2 sin θ – 3 cos θ)/ (4 sin θ – 9 cos θ)

Solution:

Given,

12 cosec θ = 13

⇒ cosec θ = 13/12

In right ∆ ABC, we have ∠B = 90^o

∠A = θ

So, cosec θ = AC/BC = 13/12

AC = 13 and BC = 12

By Pythagoras theorem,

AB = √(AC² – BC²)

AB = √[(13)² – (12)²]

AB = √(169 – 144)

AB = √25 = 5

Now,

sin θ = BC/AC = 12/13

cos θ = AB/AC = 5/13

Hence,

(2 sin θ – 3 cos θ)/ (4 sin θ – 9 cos θ)

= (2 x 12/13 – 3 x 5/13)/(4 x 12/13 – 9 x 5/13) = 3

Given: sin A = 3/5, find: (i) tan A (ii) cos A

Solution:

In right ∆ ABC, we have ∠B = 90^o

Given, sin A = 3/5

⇒ perpendicular/hypotenuse = 3/5

BC/AC = 3/5

Hence,

If the length of BC is 3x, the length of AC is 5x

We have,

AB² + BC² = AC² [By Pythagoras Theorem]

AB² + (3x)² = (5x)²

AB² = 25x² – 9x²

= 16x²

Taking square root on both sides, we get

AB = 4x, which is the base

Now,

(i) tan A = perpendicular/base

= 3x/4x

= 3/4

(ii) cos A = base/hypotenuse

= 4x/5x

= 4/5

Given: cos A = 5/13 Evaluate: (i) (sin A – cot A)/2tan A (ii) cot A + 1/cos A

Solution:

In right ∆ ABC, we have ∠B = 90^o

 Given, cos A = 5/13

⇒ base/hypotenuse = 5/13

AB/AC = 5/13

Hence,

If length of AB = 5x, the length of AC = 13x

So, by Pythagoras Theorem

AB² + BC² = AC²

(5x)² + BC² = (13x)²

BC² = 169x² – 25x²

= 144x²

Taking square root on both sides, we get

BC = 12x, which is the perpendicular

Now,

tan A = perpendicular/base

= 12x/5x

= 12/5

sin A = perpendicular/base

= 12x/13x

= 12/13

cot A = base/perpendicular

= 5x/12x

= 5/12

(i) (sin A – cot A)/2tan A = [(12/13) – (5/12)]/ 2(12/5)

= 79/156 x 5/24

= 395/3744

(ii) cot A + 1/cos A = 5/12 + 1/(5/13)

= 5/12 + 13/5

= 181/60

Given: sec A = 29/21, evaluate: sin A – 1/tan A

Solution:

In right ∆ ABC, we have ∠B = 90^o

Given, sec A = 29/21

⇒ hypotenuse/base = 29/21

AC/AB = 29/21

Hence,

If length of AB = 21x, the length of AC = 29x

So, by Pythagoras Theorem

AB² + BC² = AC²

(21x)² + BC² = (29x)²

BC² = 841x² – 441x²

= 400x²

Taking square root on both sides, we get

BC = 20x, which is the perpendicular

Now,

sin A = perpendicular/hypotenuse

= 20x/29x

= 20/29

tan A = perpendicular/base

= 20x/21x

= 20/21

Therefore,

sin A – 1/tan A = 20/29 – 1/(20/21)

= 20/29 – 21/20

= – 209/580

Given: tan A = 4/3, find: cosec A/(cot A – sec A)

Solution:

In right ∆ ABC, we have ∠B = 90^o

Given, tan A = 4/3

⇒ perpendicular/base = 4/3

BC/AB = 4/3

Hence,

If length of AB = 3x, the length of BC = 4x

So, by Pythagoras Theorem

AB² + BC² = AC²

(3x)² + (4x)² = AC²

AC² = 9x² + 16x²

= 25x²

Taking square root on both sides, we get

AC = 5x, which is the hypotenuse

Now,

sec A = hypotenuse/base

= AC/AB

= 5x/3x

= 5/3

cot A = base/perpendicular

= AB/BC

= 3x/4x

= ¾

cosec A = hypotenuse/perpendicular

= AC/BC

= 5x/4x

= 5/x

Therefore,

cosec A/(cot A – sec A) = (5/4)/(3/4 – 5/3)

= (5/4)/(-11/12)

= – 60/44

= – 15/11

Given: 4 cot A = 3, find; (i) sin A (ii) sec A (iii) cosec² A – cot²A.

Solution:

In right ∆ ABC, we have ∠B = 90^o

Given, 4 cot A = 3

cot A = 3/4

⇒ base/perpendicular = 4/3

AB/BC = 3/4

Hence,

If length of AB = 3x, the length of BC = 4x

So, by Pythagoras Theorem

AB² + BC² = AC²

(3x)² + (4x)² = AC²

AC² = 9x² + 16x²

= 25x²

Taking square root on both sides, we get

AC = 5x, which is the hypotenuse

Now,

(i) sin A = perpendicular/hypotenuse

= 4x/5x

= 4/5

(ii) sec A = hypotenuse/base

= AC/AB

= 5x/3x

= 5/3

(iii) cosec A = hypotenuse/perpendicular

= AC/BC

= 5x/4x

= 5/4

cot A = 3/4

Hence,

cosec² A – cot² A = (5/4)² – (3/4)²

= (25 – 9)/16

= 16/16

= 1

Find Sin A and Sec A, if 15 cot A = 8.

Solution:

Given that 15 cot A = 8

Therefore, cot A = 8/15.

We know that tan A = 1/ cot A

Hence, tan A = 1/(8/15) = 15/8.

Thus, Side opposite to ∠A/Side Adjacent to ∠A = 15/8

Let BC be the side opposite to ∠A and AB be the side adjacent to ∠A and AC be the hypotenuse of the right triangle ABC, respectively.

Hence, BC = 15x and AB = 8x.

Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.

(i.e) AC² = AB² + BC²

AC² = (8x)²+(15x)²

AC² = 64x²+225x²

AC² = 289x²

AC = 17x.

Therefore, the hypotenuse AC = 17x.

Finding Sin A:

We know Sin A = Side Opposite to ∠A / Hypotenuse

Sin A = 15x/17x

Sin A = 15/17.

Finding Sec A:

To find Sec A, find cos A first.

Thus, cos A = Side adjacent to ∠A / Hypotenuse

Cos A = 8x/17x

We know that sec A = 1/cos A.

So, Sec A = 1/(8x/17x)

Sec A = 17x/8x

Sec A = 17/8.

Therefore, Sin A = 15/17 and sec A = 17/8.

If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that  0° <A+B ≤ 90°; A>B ]

Solution:

Given that 

Tan (A+B) = √3.

We know that tan 60 = √3.

Thus, tan (A+B) = tan 60° = √3.

Hence A+B= 60° …(1)

Similarly, given that,

Tan (A-B) = 1/√3.

We know that tan 30° = 1/√3.

Thus, tan (A-B) = tan 30° = 1/√3.

Hence, A-B = 30° …(2)

Now, adding the equations (1) and (2), we get

A+B+A-B = 60° + 30°

2A = 90°

A = 45°.

Now, substitute A = 45° in equation (1), we get

45° +B = 60°

B = 60°- 45°

B = 15°

Hence, A = 45 and B = 15°.

💡 Do You Know?

• The values of trigonometric ratios are exact for 0°, 30°, 45°, 60°, and 90°.
• These ratios are frequently used in geometry, physics, astronomy, and engineering.
• The identities like sin²θ + cos²θ = 1 help derive all other formulas in trigonometry.

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