Introduction to Trigonometry Question - CBSE

Question

If $\sec heta + an heta = p$, prove that:
(i) $\sec heta - an heta = \dfrac{1}{p}$ [1.5 Marks]
(ii) $\sec heta = \dfrac{p^2 + 1}{2p}$ [1.5 Marks]
(iii) $\sin heta = \dfrac{p^2 - 1}{p^2 + 1}$ [2.0 Marks]

Accepted Answer

(i) $\sec^2 heta - an^2 heta = 1 \Rightarrow (\sec heta + an heta)(\sec heta - an heta) = 1 \Rightarrow \sec heta - an heta = \dfrac{1}{p}$. [1.5 Marks]
(ii) Add $\sec heta + an heta = p$ and $\sec heta - an heta = 1/p$: $2\sec heta = p + \dfrac{1}{p} = \dfrac{p^2+1}{p} \Rightarrow \sec heta = \dfrac{p^2+1}{2p}$. [1.5 Marks]
(iii) Subtract: $2 an heta = \dfrac{p^2-1}{p} \Rightarrow an heta = \dfrac{p^2-1}{2p}$. $\sin heta = \dfrac{ an heta}{\sec heta} = \dfrac{p^2-1}{p^2+1}$. Proved! [2.0 Marks]

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🎯 Official CBSE Marking Scheme:
Part (i) Proving $\sec heta - an heta = 1/p$: 1.5 Marks
Part (ii) Solving $\sec heta = (p^2+1)/(2p)$: 1.5 Marks
Part (iii) Solving $\sin heta = (p^2-1)/(p^2+1)$: 2.0 Marks

Question:

Step-by-Step Solution

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