If -tan- p- Show that p 2-1- p 2-1-sin-4 MARKS -

Concept: 2 Marks Application: 2 Marks p^2 1/p^2+1 =sec^2 θ + tan^2 θ + 2 sec θ tan θ sec^2 θ+tan^2 θ/sec^2 θ + tan^2 θ + 2 sec θ tan θ+sec^2 θ tan^2 θ =2 t ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

If θ+tanθ= p....

Question

If $s e c θ + t a n θ = p$ . Show that $\frac{p^{2} - 1}{p^{2} + 1} = s i n θ$ [4 MARKS]

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Solution

Concept: 2 Marks
Application: 2 Marks

$\frac{p^{2} - 1}{p^{2} + 1}$

$= \frac{s e c^{2} θ + t a n^{2} θ + 2 s e c θ t a n θ - s e c^{2} θ + t a n^{2} θ}{s e c^{2} θ + t a n^{2} θ + 2 s e c θ t a n θ + s e c^{2} θ - t a n^{2} θ}$

$= \frac{2 t a n^{2} θ + 2 s e c θ t a n θ}{2 s e c^{2} θ + 2 s e c θ t a n θ}$

$= \frac{2 t a n θ (t a n θ + s e c θ)}{2 s e c θ (s e c θ + t a n θ)}$

$= \frac{\frac{s i n θ}{c o s θ}}{\frac{1}{c o s θ}} = s i n θ$

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Similar questions

Q. If

s e c θ + t a n θ = p

, prove that

s i n θ = \frac{p^{2} - 1}{p^{2} + 1}

Q. If (sec θ + tan θ ) = p then show that (sec θ – tan θ) =

\frac{1}{p}

.
Hence, show that cos θ =

\frac{2 p}{(p^{2} + 1)} and \sin θ = \frac{p^{2} - 1}{p^{2} + 1}

Q. If

sec θ + tan θ = p

then

sin θ = \frac{p^{2} + 1}{p^{2} - 1}

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

Q. If

sec θ + tan θ = p

then prove that

\frac{p^{2} - 1}{p^{2} + 1} = tan θ

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If sec θ+tan θ=p. Show that p2−1p2+1=sin θ [4 MARKS]

Concept: 2 Marks Application: 2 Marks p2−1p2+1 =sec2 θ+tan2 θ+2 sec θ tan θ−sec2 θ+tan2 θsec2 θ+tan2 θ+2 sec θ tan θ+sec2 θ−tan2 θ =2 tan2 θ+2 sec θ tan θ2 sec2 θ+2 sec θ tan θ =2 tan θ(tan θ+sec θ)2 sec θ(sec θ+tan θ) =sin θcos θ1cos θ=sin θ

If $s e c θ + t a n θ = p$ . Show that $\frac{p^{2} - 1}{p^{2} + 1} = s i n θ$ [4 MARKS]