Find the value of sin 60-cos 30-

The correct option is C 0 nbsp; nbsp;sin(60^∘+θ) cos(30^∘ θ) = cos(90^∘ (60^∘ +θ)) cos(30^∘ θ) [∵ cos(90^∘ θ) = sin θ ] = cos(30^∘ θ) cos(30^∘ θ) = ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios of Standard Angles

Find the valu...

Question

Find the value of sin(60+ $θ$ ) - cos(30- $θ$ ).

2 cos $θ$

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2 sin $θ$

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Solution

The correct option is C
0

$s i n (60^{\circ} + θ) - c o s (30^{\circ} - θ)$
= $c o s (90^{\circ} - (60^{\circ} + θ)) - c o s (30^{\circ} - θ)$
[ $∵ c o s (90^{\circ} - θ) = s i n θ$ ]
= $c o s (30^{\circ} - θ) - c o s (30^{\circ} - θ)$
= 0

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Similar questions

Find the value of $s i n (60 ° + θ) - c o s (30 ° - θ) .$

Q. If

(60^{\circ} + θ)

and

(30^{\circ} - θ)

are acute angles, find the value of

s i n (60^{\circ} + θ) - c o s (30^{\circ} - θ)

Q. The value of

sin (60^{\circ} + θ) - c o s (30^{\circ} - θ)

is equal to

Q. sin (60° + θ) – cos (30° – θ) = ?
(a) 2sin θ
(b) 2cos θ
(c) 0
(d) 1

Q. Find the value of the following

(i) sin 60^{\circ} cos 30^{\circ} + cos 60^{\circ} sin 30^{\circ}

(i i) cos 30^{\circ} cos 60^{\circ} - sin 30^{\circ} sin 60^{\circ}

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Find the value of sin(60+θ) - cos(30-θ).

The correct option is C 0 sin(60∘+θ)−cos(30∘−θ) = cos(90∘−(60∘+θ))−cos(30∘−θ) [∵cos(90∘−θ)=sinθ ] = cos(30∘−θ)−cos(30∘−θ) = 0

Find the value of sin(60+ $θ$ ) - cos(30- $θ$ ).

The correct option is C
0

$s i n (60^{\circ} + θ) - c o s (30^{\circ} - θ)$
= $c o s (90^{\circ} - (60^{\circ} + θ)) - c o s (30^{\circ} - θ)$
[ $∵ c o s (90^{\circ} - θ) = s i n θ$ ]
= $c o s (30^{\circ} - θ) - c o s (30^{\circ} - θ)$
= 0