TS 10th Class Maths Solutions Chapter 11 Trigonometry Ex 11.3

Students can practice 10th Class Maths Study Material Telangana Chapter 11 Trigonometry Ex 11.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 11 Trigonometry Exercise 11.3

Question 1.
Evaluate. (AS₄)

i) \(\frac{\tan 36}{\cot 54}\)
Solution:
Given that \(\frac{\tan 36}{\cot 54}\)
= \(\frac{\tan 36}{\cot \left(90^{\circ}-36^{\circ}\right)}\) = \(\frac{\tan 36^{\circ}}{\tan 36^{\circ}}\) = 1

ii) cos 12° – sin 78°
Solution:
Given that cos 12° – sin 78°
= cos 12° – sin (90° – 12°)
= cos 12° – cos 12°
= 0 [∵ sin (90° – θ) = cos θ]

iii) cosec 31° – sec 59°
Solution:
Given that cosec 31° – sec 59°
= cosec 31° – sec (90° – 31°) [∵ sec (90°-31°) = cosec 31°]
= cosec 31° – cosec 31°
= 0

iv) sin 15° sec 75°
Solution:
Given that sin 15° sec 75°
= sin 15° sec (90° – 15°)
= sin 15° cosec 15° [∵ sec (90° – θ) = cosec θ]
1 sin 15°
= 1

v) tan 26° tan 64°
Solution:
Given that tan 26° tan 64°
= tan 26°. tan (90° – 26°)
= tan 26° . cot 26° [∵ tan (90° – θ) = cot θ]
= tan 26° . \(\frac{1}{\tan 26^{\circ}}\)
= 1

Question 2.
Show that (AS₂)
i) tan 48° tan 16° tan 42° tan 74° = 1
Solution:
L.H.S. = tan 48° tan 16° tan 42° tan 74°
= tan 48°. tan 16°. tan (90° – 48°) . tan (90° – 16°)
= tan 48°. tan 16°. cot 48°. cot 16° [∵ tan (90 – θ) = cot θ]
= tan 48° tan 16° \(\frac{1}{\tan 48^{\circ}}\) \(\frac{1}{\tan 16^{\circ}}\) [∵ cot θ = \(\frac{1}{\tan \theta}\)]
= 1 = R.H.S.

ii) cos 36° cos 54° – sin 36° sin 54° = 0
Solution:
L.H.S. = cos 36° cos 54° – sin 36° sin 54°
= cos (90° – 54°). cos (90° – 36°) – sin 36°. sin 54°
= sin 54°. sin 36° – sin 36°. sin 54° [∵ cos (90 – θ) = sin θ]
= 0 = R.H.S.
∴ L.H.S. = R.H.S.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle. Find the value of A. (AS₁, AS₄)
Solution:
Given that, tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵ tan θ = cot (90 – θ)]
⇒ 90° – 2A = A – 18°
⇒ 108° = 3A
∴ A = \(\frac{108^{\circ}}{3}\) = 36°
Hence, the value of ‘A’ is 36°

Question 4.
If tan A = cot B where A and B are acute angles, prove that A + B = 90°. (AS₂)
Solution:
Given that tan A = cot B
⇒ cot (90° – A) = cot B
[∵ tan θ = cot (90 – θ)]
⇒ 90° – A = B
∴ A + B = 90°

Question 5.
If A, B and C are interior angles of a triangle ABC, then show that tan \(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) = cot \(\left(\frac{\mathrm{C}}{2}\right)\)
Solution:
Given A, B and C are interior angles of right angle triangle ABC then
A + B + C = 180°
On dividing the above equation by ‘2’ on both sides, we get
\(\frac{\mathrm{A}+\mathrm{B}}{2}\) + \(\frac{\mathrm{C}}{2}\) = \(\frac{180^{\circ}}{2}\) = 90°
\(\frac{\mathrm{A}+\mathrm{B}}{2}\) = 90° – \(\frac{\mathrm{C}}{2}\)
On taking ‘tan’ ratio on both sides.
tan \(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) = tan \(\left(90^{\circ}-\frac{C}{2}\right)\)
tan \(\left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\) = cot \(\frac{\mathrm{C}}{2}\)
Hence proved.

Question 6.
Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°. (AS₃)
Solution:
We have sin 75° + cos 65°
= sin (90° – 15°) + cos (90° – 25°)
= cos 15° + sin 25°
[∵ sin (90 – θ) = cos θ and
cos (90 – θ) = sin θ]