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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.
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Solution
sin θ = `("AB")/("AC")` and tan θ = `("AB")/("BC")`
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Proof: L.H.S. = sec θ + tan θ
= `1/square + square/square`
= `square/square` ......`(∵ sec θ = 1/square, tan θ = square/square)`
= `((1 + sin θ) square)/(cos θ square)` ......[Multiplying `square` with the numerator and denominator]
= `(1^2 - square)/(cos θ square)`
= `square/(cos θ square)`
= `cos θ/(1 - sin θ)` = R.H.S.
∴ L.H.S. = R.H.S.
∴ sec θ + tan θ = `cos θ/(1 - sin θ)`
Prove that `tan θ/(1 - cot θ) + cot θ/(1 - tanθ)` = 1 + sec θ cosec θ
