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NARA [144]
3 years ago
12

4

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
4 0

Answer:

Step-by-step explanation:

D

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Prove this qns plzz ​
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Answer:

L.H.S.

= (cos5a.sin2a-cos4a.sin3a)/ (sin5a.sin2a-cos4a.cos3a)

Multiply numerator and denominator by 2.

= 2(cos5a.sin2a - cos4a.sin3a) / 2(sin5a.sin2a - cos4a.cos3a)

= (2cos5a.sin2a - 2cos4a.sin3a)/

(2sin5a.sin2a - 2cos4a.cos3a) = [sin(5a+2a)-sin(5a-2a)-sin(4a+3a)

+sin(4a-3a)]/[cos(5a-2a)-cos(5a+2a)-sin(4a-3a) +cos(4a+3a)]

= (sina - sin3a)/(cso3a-cosa)

= (-2cos2a.sina)/(-2sin2a.sina)

= cos2a/sin2a

= cot2a

= R.H.S.

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3 years ago
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Maria incorrectly placed the decimal point when she wrote 0.65 inch fo the width of her computer. what is the correct decimal nu
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Two similar cylinders have surface areas of 24pi cm^2 and 54pi cm^2. The volume of the smaller cylinder is 16pi cm^2. What is th
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