If secθ+tanθ=P then P2+1P2−1 is (Question:327724.0)

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Zishan Rahman

· started a discussion

· 1 Months ago

If secθ+tanθ=P then P2+1P2−1 is (Question:327724.0)

from 2nd step till last is not clear...

Question:

If $s e c θ + t a n θ = P$ then $\frac{P^{2} + 1}{P^{2} - 1}$ is

Options:

$s i n θ$

cosec$\theta$

$t a n θ$

$s e c θ$

Solution:

$\begin{matrix} P = s e c θ + t a n θ = \frac{1}{s e c θ - t a n θ} \\ P^{2} = \frac{s e c θ + t a n θ}{s e c θ - t a n θ} \\ P^{2} + 1 = \frac{2 s e c θ}{s e c θ - t a n θ} \\ P^{2} - 1 = \frac{2 t a n θ}{s e c θ - t a n θ} \\ N o w, \frac{P^{2} + 1}{P^{2} + 1} = \frac{2 s e c θ}{2 t a n θ} = c o s e c θ \end{matrix}$ 1=2secθ2tanθ=cosecθ $\begin{matrix} P = s e c θ + t a n θ = \frac{1}{s e c θ - t a n θ} \\ P^{2} = \frac{s e c θ + t a n θ}{s e c θ - t a n θ} \\ P^{2} + 1 = \frac{2 s e c θ}{s e c θ - t a n θ} \\ P^{2} - 1 = \frac{2 t a n θ}{s e c θ - t a n θ} \\ N o w, \frac{P^{2} + 1}{P^{2} + 1} = \frac{2 s e c θ}{2 t a n θ} = c o s e c θ \end{matrix}$

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If secθ+tanθ=P then P2+1P2−1 is (Question:327724.0)

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