Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.5

Question 1.

Find the value of cos 2A , A lies in the first quadrant when

(i) cos A = \(\frac{15}{17}\)

Answer:

we know sin^{2} A + cos^{2} A = 1

sin^{2} A = 1 – cos^{2} A

Since A lies in the first quadrant, sin A is positive

∴ sin A = \(\frac{8}{17}\)

cos 2A = cos^{2} A – sin^{2} A

(ii) sin A = \(\frac{4}{5}\)

Answer:

we know sin^{2} A + cos^{2} A = 1

cos^{2} A = 1 – sin^{2}A

Since A lies in the first quadrant, cos A is positive

∴ cos A = \(\frac{3}{5}\)

cos 2A = cos^{2} A – sin^{2} A

(iii) tan A = \(\frac{16}{63}\)

Answer:

Question 2.

If θ be an acute angle, find

(i) sin \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\), when sin θ = \(\frac{1}{25}\)

(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)

Answer:

(i) sin \(\left(\frac{\pi}{4}-\frac{\theta}{2}\right)\), when sin θ = \(\frac{1}{25}\)

(ii) cos \(\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\), when sin θ = \(\frac{8}{9}\)

Question 3.

If cos θ =\(\frac{1}{2}\left(a+\frac{1}{a}\right)\), show that cos 3θ = \(\frac{1}{2}\left(a^{3}+\frac{1}{a^{3}}\right)\)

Answer:

Question 4.

Prove that

cos 5θ = 16 cos^{5}θ – 20 cos^{3}θ + 5 cos θ

Answer:

Question 5.

Prove that sin 4α = 4 tan α \(\frac{1-\tan ^{2} \alpha}{\left(1+\tan ^{2} \alpha\right)^{2}}\)

Answer:

Question 6.

If A + B = 45°, show that (1 + tan A) (1 + tan B) = 2

Answer:

Given A + B = 45°

tan(A + B) = tan 45°

tan A + tan B = 1 – tan A . tan B —— (1)

(1 + tan A)(1 + tan B) = 1 + tan B + tan A + tan A tan B

= 1 + (tan A + tan B) + tan A tan B

= 1 + 1 – tan A tan B + tan A tan B (By equation (1))

= 2

Question 7.

Prove that (1 + tan 1°) (1 + tan 2°) (1 + tan 3°) …….. (1 + tan 44°) is a multiple of 4.

Answer:

1 + tan 44° = 1 + tan (45° – 1°)

(1 + tan 1°)(1 + tan 44°) = 2

Similarly (1 + tan 2°) (1 + tan 43°) = 2

(1 + tan 3°) (1 + tan 42°) = 2

(1 + tan 22°) (1 + tan 23°) = 2

= (1 + tan 1°) (1 + tan 2°)… (1 + tan 44°) = 2 × 2 × … 22 times

It is a multiple of 4.

Question 8.

Prove that tan \(\left(\frac{\pi}{4}+\theta\right)\) – tan \(\left(\frac{\pi}{4}-\theta\right)\) = 2 tan 2θ

Answer:

Question 9.

Show that

Answer:

Question 10.

prove that (1 + sec 2θ) (1 + sec 4θ) …………….. (1 + sec 2^{n}θ) = tan 2^{n}θ . cot θ.

Answer:

Question 11.

Prove that

Answer: