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Questions
If `cos θ = 12/13`, show that `sin θ (1 - tan θ) = 35/156`.
If `cos θ = 12/13`, verify that `sin θ (1 - tan θ) = 35/156`.
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Solution
Given: cos θ = `12/13`
To prove: sin θ (1 – tan θ) = `35/156`
Proof: We know, cos θ = `B/H`
where the right-angled triangle’s base is B and its hypotenuse is H. ∠ACB = 8 is achieved by building a right triangle ABC at a right angle to B.
AB is perpendicular, BC = 12 is base, and AC = 13 is hypotenuse.
According to Pythagoras theorem, we have
AC2 = AB2 + BC2
132 = AB2 + 122
169 = AB2 + 144
169 – 144 = AB2
25 = AB2
AB = `sqrt25`
AB = 5

sin θ = `P/H = 5/13`
So, tan θ = `P/H = 5/12`
Put the values in sin θ (1 – tan θ) to find its value,
sin θ (1 – tan θ) = `15/3 (1 - 5/12)`
= `5/13 xx 7/12`
= `35/156`
Hence Proved.
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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.
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Prove that: cot θ + tan θ = cosec θ·sec θ
Proof: L.H.S. = cot θ + tan θ
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= `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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