The value of sin Bcos 90-B-cos B90-B is

The correct option is B 1 sin B cos (90 B)+cos Bsin (90 B) #160; #160; = = sin B sin B + cos B cos B #160; #160; #160; = #160; sin ...

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Standard XII

Mathematics

Basic Inverse Trigonometric Functions

the value of ...

Question

the value of $sin B cos (90^{\circ} - B) + cos B (90^{\circ} - B)$ is

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sin B cos B

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2 {sin}^{2} B

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Solution

The correct option is B 1
$sin B cos (90 - B) + cos B sin (90 - B)$
= $= sin B sin B + cos B cos B$
= ${sin}^{2} B + {cos}^{2} B = 1$

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the value of sinBcos(90∘−B)+cosB(90∘−B) is

The correct option is B 1sinBcos(90−B)+cosBsin(90−B)==sinBsinB+cosBcosB= sin2B+cos2B=1

the value of $sin B cos (90^{\circ} - B) + cos B (90^{\circ} - B)$ is

The correct option is B 1
$sin B cos (90 - B) + cos B sin (90 - B)$
= $= sin B sin B + cos B cos B$
= ${sin}^{2} B + {cos}^{2} B = 1$