Question

8. cot 10° + tan 5° equal to :
(a) sec 10°
(c) cosec 5°
(b) sec 50
(d) cosec 10°
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Answer 1

Answer:

option (d) is correct

Step-by-step explanation:

$cot\,10^{\circ}+tan\,5^{\circ}$

$=\frac{cos\,10^{\circ}}{sin\,10^{\circ}}+\frac{sin\,5^{\circ}}{cos\,5^{\circ}}$

$=\frac{cos\,10^{\circ}\,cos\,5^{\circ}+sin\,10^{\circ}\,sin\,5^{\circ}}{sin\,10^{\circ}\,cos\,5^{\circ}}$

Using

$\boxed{\bf\,cos(A-B)=cosA\,cosB+sinA\,sinB}$

$=\frac{cos(10^{\circ}-5^{\circ})}{sin\,10^{\circ}\,cos\,5^{\circ}}$

$=\frac{cos\,5^{\circ}}{sin\,10^{\circ}\,cos\,5^{\circ}}$

$=\frac{1}{sin\,10^{\circ}}$

$=cosec\,10^{\circ}$

$\implies\boxed{\bf\:cot\,10^{\circ}+tan\,5^{\circ}=cosec\,10^{\circ}}$

Answer 2

Trigonometric formulae:

Before we solve this problem, let us get to know some trigonometric formulae,

1. cotθ = 1 / tanθ

2. tan2θ = 2 tanθ / (1 - tan²θ)

3. sin2θ = 2 tanθ / (1 + tan²θ)

4. 1 / sinθ = cosecθ

Answer:

Option (d), cosec10° is correct.

Solution:

Now, cot10° + tan5°

= 1/tan10° + tan5°

= 1/{2 tan5° / (1 - tan²5°)} + tan5°

= (1 - tan²5°) / (2 tan5°) + tan5°

= (1 - tan²5° + 2 tan²5°) / (2 tan5°)

= (1 + tan²5°) / (2 tan5°)

= 1/{2 tan5° / (1 + tan²5°)}

= 1 / sin10°

= cosec10°

Hence, the solution.