If sec -tan- p- Show that p2-1-p2-1-sin-

We have, LHS = p^2 1/p^2 + 1 =(sin θ + tan θ)^2 1/(sec θ + tan θ)^2 + 1 = sec^2θ + tan^2 θ + 2sec θ tan θ 1/sec^2 θ + tan^2 θ + 2 sec θ tan θ + 1 = (sec^2 θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

If sec Θ+tanΘ...

Question

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

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Solution

$We have,$
$L H S = \frac{p^{2} - 1}{p^{2} + 1} = \frac{(s i n θ + t a n θ)^{2} - 1}{(s e c θ + t a n θ)^{2} + 1}$
$= \frac{s e c^{2} θ + t a n^{2} θ + 2 s e c θ t a n θ - 1}{s e c^{2} θ + t a n^{2} θ + 2 s e c θ t a n θ + 1}$
$= \frac{(s e c^{2} - θ) + t a n^{2} θ + 2 s e c θ t a n θ}{s e c^{2} + 2 s e c θ t a n θ + (1 + t a n^{2} θ)}$
$= \frac{t a n^{2} θ + t a n^{2} θ + 2 s e c θ t a n θ}{s e c^{2} θ + 2 s e c θ t a n θ + s e c^{2} θ}$
$= \frac{2 t a n^{2} θ + 2 s e c θ t a n θ}{2 s e c^{2} θ + 2 s e c θ t a n θ}$
$= \frac{2 t a n θ (t a n θ + s e c θ)}{2 s e c θ (s e c + t a n θ)}$
$= s i n θ = R H S$

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If sec Θ+tan Θ=p. Show that p2−1p2+1=sin Θ

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