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Evaluate the Following : Cot 40°/Cos 35° - 1/2 [Cos 35°/Sin 55°] - CBSE Class 10 - Mathematics

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Question

Evaluate the following :

`(cot 40^@)/cos 35^@ -  1/2 [(cos 35^@)/(sin 55^@)]`

Solution

We have to find:

`(cot 40^@)/cos 35^@ -  1/2 [(cos 35^@)/(sin 55^@)]`

Since `cot(90^@ - theta) = tan theta` and `cos (90^@- theta) = sin theta`

`(cot 40^@)/(tan 50^@) - 1/2 ((cos 35^@)/(sin 55^@)) = cot(90^@ - 50^@)/ tan 50^@  - 1/2 (cos(90^@ - 55^@)/sin 55^@)`

`= tan 50^@/tan 50^@  - 1/2 ((sin 55^@)/(sin 55^@))`

`= 1 - 1/2`

`= 1/2`

So value of `(cot 40^@)/(tan 50^@) - 1/2 ((cos 35^@)/(sin 55^@)) " is " 1/2`

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Solution Evaluate the Following : Cot 40°/Cos 35° - 1/2 [Cos 35°/Sin 55°] Concept: Trigonometric Ratios of Some Specific Angles.
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