AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 11 Trigonometry Ex 11.4 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 11th Lesson Trigonometry Exercise 11.4

### 10th Class Maths 11th Lesson Trigonometry Ex 11.4 Textbook Questions and Answers

Question 1.

Evaluate the following:

i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Answer:

Given (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

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

Answer:

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

= (sin^{2} θ + cos^{2} θ + 2 sin θ cos θ) + (sin^{2} θ + cos^{2} θ – 2 sin θ cos θ) [∵ (a + b)^{2} = a^{2} + b^{2} + 2ab

(a – b)^{2} = a^{2} + b^{2} – 2ab]

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

= 1 + 1

= 2

iii) (sec^{2} θ – 1) (cosec^{2} θ – 1)

Answer:

Given (sec^{2} θ – 1) (cosec^{2} θ – 1)

= tan^{2} θ × cot^{2} θ [∵ sec^{2} θ – tan^{2} θ = 1; cosec^{2} θ – cot^{2} θ = 1]

= tan^{2} θ × \(\frac{1}{\tan ^{2} \theta}\) = 1

Question 2.

Show that (cosec θ – cot θ)^{2} = \(\frac{1-\cos \theta}{1+\cos \theta}\)

Answer:

Question 3.

Show that \(\sqrt{\frac{1+\sin A}{1-\sin A}}\) = sec A + tan A

Answer:

Given that L.H.S. = \(\sqrt{\frac{1+\sin A}{1-\sin A}}\)

Rationalise the denominator, rational factor of 1 – sin A is 1 + sin A.

[∵ (a + b)(a + b) = (a + b)^{2}]; (a – b)(a + b) = a^{2} — b^{2}]

= \(\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}\)

= \(\frac{1+\sin A}{\cos A}\)

= \(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\)

= sec A + tan A = R.H.S.

Question 4.

Show that \(\frac{1-\tan ^{2} A}{\cot ^{2} A-1}\) = tan^{2} A

Answer:

Question 5.

Show that \(\frac{1}{\cos \theta}\) – cos θ = tan θ – sin θ.

Answer:

Question 6.

Simplify sec A (1 – sin A) (sec A + tan A)

Answer:

L.H.S. = sec A (1 – sin A) (sec A + tan A)

= (sec A – sec A . sin A) (sec A + tan A)

= (sec A – \(\frac{1}{\cos A}\) . sin A) (sec A + tan A)

= (sec A – tan A) (sec A + tan A)

= sec^{2} A – tan^{2} A [∵ sec^{2} A – tan^{2} A = 1]

= 1

Question 7.

Prove that (sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7 + tan^{2} A + cot^{2} A

Answer:

L.H.S. = (sin A + cosec A)^{2} + (cos A + sec A)^{2}

= (sin^{2} A + cosec^{2} A + 2 sin A . cosec A) + (cos^{2} A – sec^{2} A + 2 cos A . sec A) [∵ (a + b)^{2} = a^{2} + b^{2} + 2ab]

= (sin^{2} A + cos^{2} A) + cosec^{2} A + 2 sin A . \(\frac{1}{\sin A}\) + sec^{2} A + 2 cos A . \(\frac{1}{\cos A}\)

[∵ \(\frac{1}{\sin A}\) = cosec A; \(\frac{1}{\cos A}\) = sec A]

= 1 +(1 + cot^{2} A) + 2 + (1 + tan^{2} A) + 2

[∵ sin^{2} A + cos^{2} A = 1; cosec^{2} A = 1 + cot^{2} A; sec^{2} A = 1 + tan^{2} A]

= 7 + tan^{2} A + cot^{2} A

= R.H.S.

Question 8.

Simplify (1 – cos θ) (1 + cos θ) (1 + cot^{2} θ)

Answer:

Given that

(1 – cos θ) (1 + cos θ) (1 + cot^{2} θ)

= (1 – cos^{2} θ) (1 + cot^{2} θ)

[∵ (a – b) (a + b) = a^{2} – b^{2}]

= sin^{2} θ. cosec^{2} θ [∵ 1 – cos^{2} θ = sin^{2} θ; 1 + cot^{2} θ = cosec^{2} θ]

= sin^{2} θ . \(\frac{1}{\sin ^{2} \theta}\) [∵ cosec θ = \(\frac{1}{\sin \theta}\)]

= 1

Question 9.

If sec θ + tan θ = p, then what is the value of sec θ – tan θ?

Answer:

Given that sec θ + tan θ = p ,

We know that sec^{2} θ – tan^{2} θ = 1

sec^{2} θ – tan^{2} θ = (sec θ + tan θ) (sec θ – tan θ)

= p (sec θ – tan θ)

= 1 (from given)

⇒ sec θ – tan θ = \(\frac{1}{p}\)

Question 10.

If cosec θ + cot θ = k, then prove that cos θ = \(\frac{k^{2}-1}{k^{2}+1}\)

Answer:

Method-I:

Given that cosec θ + cot θ = k

R.H.S. = \(\frac{k^{2}-1}{k^{2}+1}\)

Method – II:

Given that cosec θ + cot θ = k ……..(1)

We know that cosec^{2} θ – cot^{2} θ = 1

⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1 [∵ a^{2} – b^{2} = (a -b)(a + b)]

⇒ k (cosec θ – cot θ) = 1

⇒ (cosec θ – cot θ) = \(\frac{1}{k}\)

By solving (1) and (2)

According to identity cos^{2} θ + sin^{2} θ = 1

Hence proved.