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Question
The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is
Options
1
2
4
0
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Solution
The given expression is
`(1+cot θ-cosec θ )(1+tan θ+sec θ)`
Simplifying the given expression, we have
`(1+cot θ-cosec θ)(1+tan θ+sec θ)`
=`(1+cos θ/sin θ-1/sin θ)(1+sin θ/cos θ+1/cos θ)`
= `(sin θ+cos θ-1)/sin θxx (cos θ+sin θ+1)/cos θ`
= `((sin θ+cos θ-1)(cos θ+sin θ+1))/(sin θcos θ)`
=`({(sin θ+cos θ)-1}{(sin θ+cos θ)+1})/(sin θ cos θ)`
=`((sin θ+cos θ)^2-(1)^2)/(sin θ cos θ)`
=`((sin θ+cos θ)^2-(1)^2)/(sin θ cos θ)`
=`((sin^2 θ+cos^2θ+2 sin θcos θ)-1)/(sin θ cos θ)`
=`((sin ^2θ+cos^2θ)+2 sinθ cos θ-1)/(sin θcos θ)`
= `(1+2 sin θ cosθ-1)/(sinθ cos θ)`
=`( 2 sin θ cos θ)/(sin θ cos θ)`
=`2`
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RELATED QUESTIONS
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If `( cosec theta + cot theta ) =m and ( cosec theta - cot theta ) = n, ` show that mn = 1.
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The value of the expression \[\sin {80}^° - \cos {80}^°\]
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
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Prove that `(tan θ)/(cot(90° - θ)) + (sec (90° - θ) sin (90° - θ))/(cosθ. cosec θ) = 2`.
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tan θ cosec2 θ – tan θ is equal to
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
(sec θ + tan θ) . (sec θ – tan θ) = ?
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S = `square`
= `square/sintheta + sintheta/costheta`
= `(cos^2theta + sin^2theta)/square`
= `1/(sintheta*costheta)` ......`[cos^2theta + sin^2theta = square]`
= `1/sintheta xx 1/square`
= `square`
= R.H.S
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
