**Chapter 6 Trigonometry Set 6.1**

## Chapter 6 Trigonometry Set 6.1

**Question 1.**

We know that,

sin

^{2}θ + cos

^{2}θ = 1

LetAB = 7k and AC = 25k

In ∆ABC, ∠B = 90°

∴ AB

^{2}+ BC

^{2}= AC

^{2}… [Pythagoras theorem]

∴ (7k)

^{2}+ BC

^{2}= (25k)

^{2}

∴ 49k

^{2}+ BC

^{2}= 625k

^{2}

∴ BC

^{2}= 625k

^{2}– 49k

^{2}

∴ BC

^{2}= 576k

^{2}

∴ BC = 24k …[Taking square root of both sides]

**Question 2.If tan θ = 3/4, find the values of sec θ and cos θ.Solution:**

Let AB = 3k and BC 4k

In ∆ABC,∠B = 90°

∴ AB

^{2}+ BC

^{2}= AC

^{2}…[Pythagoras theorem]

∴ (3k)

^{2}+ (4k)

^{2}= AC

^{2}

∴ 9k

^{2}+ 16k

^{2}= AC

^{2}

∴ AC

^{2}= 25k

^{2}

∴ AC = 5k …[Taking square root of both sides]

**Question 3.If cot θ = 40/9, find the values of cosec θ and sin θSolution:**

..[Taking square root of both sides]

Let BC = 40k and AB = 9k

In ∆ABC, ∠B = 90°

∴ AB

^{2}+ BC

^{2}= AC

^{2}… [Pythagoras theorem]

∴ (9k)

^{2}+ (40k)

^{2}= AC

^{2}

∴ 81k

^{2}+ 1600k

^{2}= AC

^{2}

∴ AC

^{2}= 1681k

^{2}

∴ AC = 41k … [Taking square root of both sides]

**Question 4.If 5 sec θ – 12 cosec θ = θ, find the values of sec θ, cos θ and sin θ.Solution:**

5 sec θ – 12 cosec θ = 0 …[Given]

∴ 5 sec θ = 12 cosec θ

**Question 5.If tan θ = 1, then find the value ofSolution:**

tan θ = 1 … [Given]

We know that, tan 45° = 1

∴ tan θ = tan 45°

∴ θ = 45°

**Question 6.**

Proof:

i. L.H.S. = sin2θcosθ+cosθ

ii. L.H.S. = cos^{2} θ(1 + tan^{2} θ)

= cos^{2} θ sec^{2} θ …[∵ 1 + tan^{2} θ = sec^{2} θ]

= 1

= R.H.S.

∴ cos^{2} θ (1 + tan^{2} θ) = 1

iv. L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ. sec θ

v. L.H.S. = cot θ + tan θ

∴ cot θ + tan θ = cosec θ.sec θ

vii. L.H.S. = sin^{4} θ – cos^{4} θ

= (sin^{2} θ)^{2} – (cos^{2} θ)^{2}

= (sin^{2} θ + cos^{2} θ) (sin^{2} θ – cos^{2} θ)

= (1) (sin^{2} θ – cos^{2} θ) ….[∵ sin^{2} θ + cos^{2} θ = 1]

= sin^{2} θ – cos^{2} θ

= (1 – cos^{2} θ) – cos^{2} θ …[θ sin^{2} θ = 1 – cos^{2} θ]

= 1 – 2 cos^{2} θ

= R.H.S.

∴ sin^{4} θ – cos^{4} θ = 1 – 2 cos^{2} θ

viii. L.H.S. = sec θ + tan θ

xi. L.H.S. = sec^{4} A (1 – sin^{4} A) – 2 tan^{2} A

= sec^{4} A [1^{2} – (sin^{2} A)^{2}] – 2 tan^{2} A

= sec^{4} A (1 – sin^{2}A) (1 + sin^{2} A) – 2 tan^{2} A

= sec^{4} A cos^{2}A (1 + sin^{2} A) – 2 tan^{2}A

[ ∵ sin^{2} θ + cos^{2} θ = 1 ,∵ 1 – sin^{2} θ = cos^{2} θ]

**Intext Questions and Activities**

**Question 1.Fill in the blanks with reference to the figure given below. (Textbook pg. no. 124)Solution:**

**Question 2.Complete the relations in ratios given below. (Textbook pg, no. 124)Solution:**

ii. sin θ = cos (90 – θ)

iii. cos θ = (90 – θ)

iv. tan θ × tan (90 – θ) = 1

**Question 3.Complete the equation. (Textbook pg. no, 124)sin**

^{2}θ + cos

^{2}θ = [______]Solution:

sin

^{2}θ + cos

^{2}θ = [1]

**Question 4.Write the values of the following trigonometric ratios. (Textbook pg. no. 124)Solution:**