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Definition of Functions

If sec Θ + ...

Question

If $s e c Θ + t a n Θ = p,$ prove that $s i n Θ = \frac{p^{2} - 1}{p^{2} + 1}$

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Solution

RHS $= \frac{P^{2} - 1}{P^{2} + 1}$
$= \frac{(s e c θ + t a n θ)^{2} - 1}{(s e c θ + t a n θ)^{2} + 1}$
$= \frac{s e c^{2} θ + t a n^{2} θ + 2 s e c θ . t a n θ - 1}{s e c^{2} θ + t a n^{2} θ + 2 s e c θ . t a n θ + 1}$
$= \frac{t a n^{2} θ + t a n^{2} θ + 2 s e c θ . t a n θ}{s e c^{2} θ + s e c^{2} θ + 2 s e c θ . t a n θ}$
$= \frac{2 t a n^{2} θ + 2 s e c θ . t a n θ}{2 s e c^{2} θ + 2 s e c θ . t a n θ}$
$= \frac{2 t a n θ (t a n θ + s e c θ)}{2 s e c θ (s e c θ + t a n θ)}$
$= \frac{t a n θ}{s e c θ} = s i n θ = L H S$

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Similar questions

(sec θ + tan θ) = p

sin θ = \frac{p^{2} - 1}{p^{2} + 1}

sec θ + tan θ = p

\frac{p^{2} - 1}{p^{2} + 1} = sin θ

If $s e c θ + t a n θ = p$ , prove that

(i) $s e c θ = \frac{1}{2} (p + \frac{1}{p})$

(ii) $t a n θ = \frac{1}{2} (p - \frac{1}{p})$

(iii) $s i n θ = \frac{p^{2} - 1}{p^{2} + 1}$

sec θ + tan θ = p

then prove that

\frac{p^{2} - 1}{p^{2} + 1} = tan θ

sec θ + tan θ = p

sin θ = \frac{p^{2} + 1}{p^{2} - 1}

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