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Question
If 3 cot θ = 2, find the value of `(4sin theta - 3 cos theta)/(2 sin theta + 6cos theta)`.
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Solution
`cot theta = 2/3`
`= ((4 sin theta - 3 cos theta)/sin theta)/((2sin theta + 6 cos theta)/sin theta)`
`= (4 - 3 cot theta)/(2 + 6 cot theta)`
`= (4 - 3 xx 2/3)/(2 + 6 xx 2/3)`
`= (4 + 2)/(2 + 4) = 2/6`
`= 1/3`
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Prove that: cot θ + tan θ = cosec θ·sec θ
Proof: L.H.S. = cot θ + tan θ
= `square/square + square/square` ......`[∵ cot θ = square/square, tan θ = square/square]`
= `(square + square)/(square xx square)` .....`[∵ square + square = 1]`
= `1/(square xx square)`
= `1/square xx 1/square`
= cosec θ·sec θ ......`[∵ "cosec" θ = 1/square, sec θ = 1/square]`
= R.H.S.
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