Without trigonometric tables, prove that:
$sin 53^{o} cos 37^{o} + cos 53^{o} sin 37^{o} = 1$

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Without trigonometric tables, prove that:

sin 53^{o} cos 37^{o} + cos 53^{o} sin 37^{o} = 1

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Text solutionVerified

L H S = sin 53^{o} cos 37^{o} + cos 53^{o} sin 37^{o} = sin 53^{o} cos (90 - 53)^{o} + cos 53^{o} sin (90 - 53)^{o}

= sin 53^{o} \times sin 53^{o} + cos 53^{o} \times cos 53^{o} = sin^{2} 53^{o} + cos^{2} 53^{o} = 1 = R H S

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Question Text	Without trigonometric tables, prove that: $sin 53^{o} cos 37^{o} + cos 53^{o} sin 37^{o} = 1$
Answer Type	Text solution:1
Upvotes	150