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Question
In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.
Solution
sin θ = `("AB")/("AC")` and tan θ = `("AB")/("BC")`
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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.
∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec·sec θ
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