The value of 60-sin40- cos10-cos40- sin10- is

60^∘[sin(40^∘+θ)cos(10^∘+θ) cos(40^∘+θ)sin(10^∘+θ)] Let A=40^∘+θ, B=10^∘+θ = 60^∘[sin A cos B cos A sin B ] =60^∘[sin(A B)] =2sin(A B) Putting the values of ...

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Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

The value of ...

Question

The value of $sec 60^{\circ} [sin (40^{\circ} + θ) cos (10^{\circ} + θ) - cos (40^{\circ} + θ) sin (10^{\circ} + θ)]$ is

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Solution

$sec 60^{\circ} [sin (40^{\circ} + θ) cos (10^{\circ} + θ) - cos (40^{\circ} + θ) sin (10^{\circ} + θ)]$
Let $A = 40^{\circ} + θ, B = 10^{\circ} + θ$
$= sec 60^{\circ} [sin A cos B - cos A sin B] = sec 60^{\circ} [sin (A - B)] = 2 sin (A - B)$
Putting the values of $A$ and $B$ ,
$= 2 sin (40^{\circ} + θ - (10^{\circ} + θ)) = 2 sin 30^{\circ} = 1$

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Prove that sin $(40^{\circ} + θ)$ cos $(10^{\circ} + θ)$ - cos $(40^{\circ} + θ)$ sin $(10^{\circ} + θ)$ = $\frac{1}{2}$

Q. The value of

\frac{\sin 70 ° + \cos 40 °}{\cos 70 ° + \sin 40 °}

is _________.

cos (40^{\circ} + θ) - sin (50^{\circ} - θ) =

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The value of sec60∘[sin(40∘+θ)cos(10∘+θ)−cos(40∘+θ)sin(10∘+θ)] is

sec60∘[sin(40∘+θ)cos(10∘+θ)−cos(40∘+θ)sin(10∘+θ)] Let A=40∘+θ, B=10∘+θ =sec60∘[sinAcosB−cosAsinB]=sec60∘[sin(A−B)]=2sin(A−B) Putting the values of A and B, =2sin(40∘+θ−(10∘+θ))=2sin30∘=1

The value of $sec 60^{\circ} [sin (40^{\circ} + θ) cos (10^{\circ} + θ) - cos (40^{\circ} + θ) sin (10^{\circ} + θ)]$ is