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Question
To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
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Solution
L.H.S. = \[\boxed{\text{cot} \phantom{.} θ + \text{tan} \phantom{.}θ}\]
= \[\frac{\boxed{\text{cos}\phantom{.}θ}}{\text{sin}\phantom{.}θ} + \frac{\text{sin}\phantom{.}θ}{\text{cos}\phantom{.}θ}\]
= \[\frac{\text{cos}^2θ + \text{sin}^2θ}{\boxed{\text{sin}θ.\text{cos}θ}}\]
= `1/(sinθ.cosθ)` ...[cos2θ + sin2θ = \[\boxed{1}\]]
= \[\frac{1}{\text{sin}θ} \times \frac{1}{\boxed{\text{cos}θ}}\]
= \[\boxed{\text{cosec} \phantom{.}θ \times \text{sec} \phantom{.}θ}\]
= R.H.S.
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