The value of cos-4- A cos-4- B -sin-4- A sin-4- B A- cos A - B B- cos A-BC- sin A - B D- sin B - A

The correct option is D sin (B A)cos(π4+A)cos(π4 B) sin(π4+A)sin(π4 B) Assuming π4+A =α, π4 B = β Now, cos(π4+A)cos(π4 B) sin(π4+A)sin(π4 B) nbsp;= nbsp ...

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Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

The value of ...

Question

The value of $cos (\frac{π}{4} + A) cos (\frac{π}{4} - B) - sin (\frac{π}{4} + A) sin (\frac{π}{4} - B)$ is

cos (A + B)

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cos (A - B)

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sin (A + B)

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sin (B - A)

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Solution

The correct option is D $sin (B - A)$
$cos (\frac{π}{4} + A) cos (\frac{π}{4} - B) - sin (\frac{π}{4} + A) sin (\frac{π}{4} - B)$
Assuming
$\frac{π}{4} + A = α, \frac{π}{4} - B = β$
Now,
$cos (\frac{π}{4} + A) cos (\frac{π}{4} - B) - sin (\frac{π}{4} + A) sin (\frac{π}{4} - B) = cos α cos β - sin α sin β = cos (α + β) = cos (\frac{π}{4} + A + \frac{π}{4} - B) = cos (\frac{π}{2} - (B - A)) = sin (B - A)$

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The value of cos(π4+A)cos(π4−B)−sin(π4+A)sin(π4−B) is

The correct option is D sin(B−A)cos(π4+A)cos(π4−B)−sin(π4+A)sin(π4−B) Assuming π4+A=α, π4−B=β Now, cos(π4+A)cos(π4−B)−sin(π4+A)sin(π4−B)=cosαcosβ−sinαsinβ=cos(α+β)=cos(π4+A+π4−B)=cos(π2−(B−A))=sin(B−A)

The value of $cos (\frac{π}{4} + A) cos (\frac{π}{4} - B) - sin (\frac{π}{4} + A) sin (\frac{π}{4} - B)$ is