Evaluate:
(i) sin 18°/cos 72° (ii) tan 26°/cot 64°
(iii) cos 48° - sin 42°(iv) cosec 31° - sec 59°
Solution:
We will be using the trigonometric ratios of complementary angles to solve the given question.
sin (90° - θ) = cos θ
tan (90° - θ) = cot θ
sec (90° - θ) = cosec θ
(i) sin 18°/cos 72°
Since, sin (90° - θ) = cos θ
Here, θ = 72°
sin 18°/cos 72° = sin (90° - 72°)/cos 72°
= cos 72° / cos 72°
= 1
(ii) tan 26°/cot 64°
tan (90° - θ) = cot θ
Here, θ = 64°
tan 26°/cot 64° = tan (90° - 64°) / cot 64°
= cot 64°/cot 64°
= 1
(iii) cos 48° - sin 42°
Since, sin (90° - θ) = cos θ
Here, θ = 48°
cos 48° - sin 42° = cos 48° - sin (90° - 48°)
= cos 48° - cos 48°
= 0
(iv) cosec 31° - sec 59°
sec (90° - θ) = cosec θ
Here, θ = 31°
cosec 31° - sec 59° = cosec 31° - sec (90° - 31°)
= cosec 31° - cosec 31°
= 0
☛ Check: NCERT Solutions for Class 10 Maths Chapter 8
Video Solution:
Evaluate: (i) sin 18°/cos 72° (ii) tan 26°/cot 64° (iii) cos 48° - sin 42° (iv) cosec 31° - sec 59°
Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.3 Question 1
Summary:
The values are as follows: (i) sin 18°/cos 72° = 1, (ii) tan 26°/cot 64° = 1, (iii) cos 48° − sin 42° = 0, (iv) cosec 31° − sec 59° = 0
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