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Question
If `tan theta = 1/sqrt7` `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + sec^2 theta) = 3/4`
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Solution
`tan theta = 1/sqrt7` `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + sec^2 theta) = 3/4`
`tan theta = "๐๐๐๐๐ ๐๐ก๐ ๐ ๐๐๐"/"๐๐๐๐๐๐๐๐ก ๐ ๐๐๐"`
Let ‘x’ be the hypotenuse
By applying Pythagoras
๐ด๐ถ2 = ๐ด๐ต2 + ๐ต๐ถ2
`x^2 = 1^2 + (sqrt7)^2`
๐ฅ2 = 1 + 7 = 8
`x = 2sqrt2`
`cosec theta = (AC)/(AB) = 2sqrt2`
`sec theta = (AC)/(BC) = (2sqrt2)/sqrt7`
Substitute, cosec θ, sec θ in equation
`=> ((2sqrt2)^2 - (2 sqrt(2/7))^2)/((2sqrt2)^2 + ((2sqrt2)/sqrt7)^2)`
`(8 - 4 xx 2/7)/(8 + 4 xx 2/7)`
`=> (8 - 8/7)/(8 + 8/7)`
`=> ((56 - 8)/7)/((56 + 8)/7)`
`=48/64`
`= 3/4`
L.H.S = R.H.S
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Proof: L.H.S. = sec θ + tan θ
= `1/square + square/square`
= `square/square` ......`(โต 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)`
What will be the value of sin 45° + `1/sqrt(2)`?
Prove that: cot θ + tan θ = cosec θ·sec θ
Proof: L.H.S. = cot θ + tan θ
= `square/square + square/square` ......`[โต cot θ = square/square, tan θ = square/square]`
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= `1/(square xx square)`
= `1/square xx 1/square`
= cosec θ·sec θ ......`[โต "cosec" θ = 1/square, sec θ = 1/square]`
= R.H.S.
∴ L.H.S. = R.H.S.
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