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प्रश्न
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
उत्तर
Given:
`cos θ=3/4`
⇒ `1/cosec θ=4/3`
⇒` sec θ=4/3`
We know that,
`sec^2θ-tan ^2 θ=1`
⇒` (4/3)^2-tan ^2 θ=1`
⇒` tan^2 θ=16/9-1`
⇒` tan^2 θ=7/9`
Therefore,
`9 tan ^2 θ+9=9xx7/9+9`
`= 7+9`
`=16`
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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
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∴ cotθ + tanθ = cosecθ × secθ
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