If -tan- p- prove that sin- p 2-1- p 2-1

R.H.S = p^2 1/p^2+1 =(secθ+tanθ)^2 1/(secθ+tanθ)^2+1 =sec^2θ+tan^2θ+2secθ tanθ 1/sec^2θ+tan^2+2secθ tanθ+1 [By (a+b)^2=a^2+b^2+2ab] =(sec^2θ 1)+tan^2θ+2secθ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identity- 2

If θ+tanθ= p,...

Question

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

Open in App

Solution

R.H.S = $\frac{p^{2} - 1}{p^{2} + 1}$

$= \frac{(s e c θ + t a n θ)^{2} - 1}{(s e c θ + t a n θ)^{2} + 1}$

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

[By $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ ]

$= \frac{(s e c^{2} θ - 1) + t a n^{2} θ + 2 s e c θ t a n θ}{s e c^{2} θ + (1 + t a n^{2} θ) + 2 s e c θ t a n θ}$

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

$[\begin{matrix} s e c^{2} θ - 1 = t a n^{2} θ s e c^{2} θ = 1 + t a n^{2} θ \end{matrix}]$

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

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

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

$= s i n θ$ = L.H.S

Suggest Corrections

Similar questions

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

s e c θ + t a n θ = p

. Show that

\frac{p^{2} - 1}{p^{2} + 1} = s i n θ

[4 MARKS]

Q. If

sec θ + tan θ = p

then prove that

\frac{p^{2} - 1}{p^{2} + 1} = tan θ

Q. If

sec θ + tan θ = p

then

sin θ = \frac{p^{2} + 1}{p^{2} - 1}

If secθ+tanθ=p ,prove that = (p²-1)/(p²+1).

Join BYJU'S Learning Program

If secθ+tanθ=p, prove that sinθ=p2−1p2+1

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