Question

Question.
$If \: sec \: \theta \: + \: tan \theta = p, show \: that$
$\frac{p {}^{2} - 1}{p {}^{2} + 1} \: cosec \: \theta = 1.$

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Answer 1

Step-by-step explanation:

LHS

(p² - 1) / (p² + 1)

{(secѲ + tanѲ)² - 1 } / {(secѲ + tanѲ)² + 1 } × cosecѲ

{sec²Ѳ + tan²Ѳ + 2secѲtanѲ - 1} / {sec²Ѳ + tan²Ѳ + 2secѲtanѲ + 1} × cosecѲ

{(sec²Ѳ - 1) + tan²Ѳ + 2secѲtanѲ} / {(tan²Ѳ + 1) + sec²Ѳ + 2secѲtanѲ} × cosecѲ

{tan²Ѳ + tan²Ѳ + 2secѲtanѲ} / {sec²Ѳ + sec²Ѳ + 2secѲtanѲ} × cosecѲ

{2tan²Ѳ + 2secѲtanѲ} / {2sec²Ѳ + 2secѲtanѲ} × cosecѲ

{2tanѲ (tanѲ + secѲ)} / {2secѲ (tanѲ + secѲ)} × cosecѲ

Now 2 and (tanѲ + secѲ) cancel out

So we have

tanѲ / secѲ × cosecѲ

sinѲ × cosecѲ

1

Therefore LHS = RHS

Answer 2

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$\rm \: If \: sec \: \theta + \tan \: \theta = p,$

$\rm \: Show \: that$

$\rm \: \: \: \: \: \frac{p {}^{2} - 1}{p {}^{2} + 1} \: cosec \: \theta = 1.$

$\huge {\red{\boxed{ \overline{ \underline{ \mid\mathfrak{\blue{SO}}{\mathrm{\green{LU}}{ \sf{\pink{TI}}}{\mathtt{\purple{ON}} \: \gray{\colon - }{\mid}}}}}}}}$

$\huge \bf \red{G} \blue{I} \pink{V} \purple{E} \orange{N},$

$\rm \: \: \: \: \: \: \:sec \: \theta + tan \theta = p \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...( \sf \red{1})$

$\bf \: We \: know \: that,$

$\rm \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: sec {}^{2} \: \theta - \tan {}^{2} \: \theta = 1$

$\implies \: \: \: \: \: \: \rm \: (sec \: \theta + tan \: \theta)(sec \: \theta - tan \: \theta) = 1$

$\implies \: \: \: \: \rm \: p(sec \: \theta - tan \: \theta) = 1 \: \: \: \: \: \: \: (by \: ( \bf \red{1}))$

$\implies \: \: \: \: \: \: \: \: \rm \: sec \: \theta - tan \: \theta = \frac{1}{p} \: \: \: \: \: \: \: \: ...( \bf \red{2 })$

$\rm \: By \: Adding \: ( \bf \red{1}) \rm \: and \: ( \bf \red{2}) \:$

$\rm \: Then, \: we \: get,$

$\rm \: \: \: 2 \: sec \: \theta = \: p + \frac{1}{p} = \frac{p {}^{2} + 1}{p}$

$\therefore \: \: \: \rm \: \: \: \: \: \: sec \: \theta \: = \frac{p {}^{2} + 1}{2p}$

$\implies \: \: \: \: \: \: \: \: \: \rm \: cos \: \theta = \frac{2p}{p {}^{2} + 1} \: \: \: \: \: \: \: \: \: \: \: \: \: ...( \bf \red{3})$

$\rm \: Eqn. \: ( \bf \red{1}) - \rm \: Eqn. \: ( \bf \red{2}) \rm \: gives$

$\: \: \: \: \: \: \: \rm2 \: tan \: \theta \: = p - \frac{1}{p} = \frac{p {}^{2} - 1}{p}$

$\therefore \: \: \: \: \: \rm \: tan \: \theta = \frac{p {}^{2} - 1}{2p} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...( \bf \red{4})$

$\rm \: Multiplying \: ( \bf \red{3}) \rm \: and \: ( \bf \red{4})$

$\rm \: We \: have,$

$\rm \: cos \: \theta. \: tan \: \theta = \frac{2p}{p {}^{2} + 1}. \frac{p {}^{2} - 1}{2p} = \frac{p {}^{2} - 1}{p {}^{2} + 1}$

$\implies \: \: \: \: \: \: \rm \: cos \: \theta. \frac{sin \: \theta}{cos \: \theta} = \frac{p {}^{2} - 1}{p {}^{2} + 1}$

$\implies \: \: \: \: \: \rm sin \ \theta = \frac{p {}^{2} - 1 }{p {}^{2} + 1}$

$\implies \: \: \: \: \rm \frac{1}{cosec \: \theta} = \frac{p {}^{2} - 1}{p {}^{2} + 1}$

$\rm \:By \: Cross - multiplying \:$

$\: \: \: \: \rm(p {}^{2} - 1) \: cosec \: \theta = p {}^{2} + 1$

$\implies \: \: \: \: \: \frac{p {}^{2} - 1}{p {}^{2} + 1} \: cosec \: \theta = 1$