Question

Prove that

sin²θ tanθ +cos²θ cotθ + 2 sinθ cosθ = tan θ+ cotθ = secθ cosecθ.

please answer fast ​

Answer 1

$\huge{\boxed{\bf{\red{Solution:-}}}}$

First member = sec ² θ tan θ + cos² θ + 2 sin θ

cos θ

$\: \: \: \: \: \: \: \: \: \sf = \: \: \: \: sin {}^{2} \: \theta \frac{sin \: \theta}{cos \: \theta} + cos {}^{2} \theta. \frac{cos \: \theta}{sin \: \theta} + 2 \: sin \: \theta \: cos \: \theta = \frac{sin {}^{3} \theta}{cos \: \theta} + \frac{cos {}^{3} \theta}{sin \: \theta} + 2 \: sin \: \theta \: cos \: \theta$

$\sf \: \: \: \: \: \: \: \: \: \: = \: \: \: \frac{sin {}^{4} \theta + cos {}^{4} \: \theta + 2 \: sin {}^{2} \: \theta \: cos {}^{2} \: \theta}{sin \: \theta \: cos \: \theta} = \frac{(sin {}^{2} \: \theta) {}^{2} + (cos {}^{2} \: \theta) {}^{2} + 2 \: sin {}^{2} \: \theta \: cos {}^{2} \: \theta}{sin \: \theta \: cos \: \theta}$

$\: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{(sin {}^{2} \theta + cos {}^{2} \: \theta) {}^{2} }{sin \: \theta \:cos \:\theta} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

[ ∵ a²+b²+2ab = (a+b)²]

$\: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \: \frac{1 {}^{2} }{sin \: \theta \: cos \: \theta} = \frac{1}{cos \: \theta}. \frac{1}{sin \: \theta}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: = \: sec \: \theta \: cosec \: \theta \: = third \: member \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(1)$

$\: \: \: \: \: \sf \: Second \: member \: = tan \: \theta + cot \: \theta$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \: \: \frac{sin \: \theta}{cos \: \theta} + \frac{cos \: \theta}{sin \: \theta} = \frac{sin {}^{2} \: \theta + cos {}^{2} \: \theta}{sin \: \theta \: cos \: \theta}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = \: \frac{1}{ sin \: \theta \: cos \: \theta} = \frac{1}{cos \: \theta}. \frac{1}{sin \: \theta}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \: sec \: \theta \: cosec \: \theta \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(2)$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \: Third \: member.$

$\implies \: \: \: \: \sf \: From \: (1) \: and \: (2), \: we \: get$

$\sf \: sin {}^{2} \: \theta \: tan \: \theta \: + \: cos {}^{2} \: \theta \: cot \: \theta \: + 2 \: sin \: \theta \: cos \: \theta = tan \: \theta \: + cot \: \theta \: = sec \: \theta \: cosec \: \theta$

Answer 2

$\huge\star\frak{\underline{AnSwer:-}}$

$\underline{\bigstar\:\textsf{Given \: in \: question:-}}$

$\bullet$ Here, we are going to solve question in two parts and we can break this in three parts after every symbol of equal.

$\bullet$So, according to this we have to prove , 1 = 2 and 2 =3,then, ultimately we will get 1 = 2 = 3.

i.e

ᴄᴀsᴇ 1:

$\normalsize\star\:\sf \red{sin^2\theta Tan\theta + cos^2\theta cot\theta + 2sin\theta cos\theta} = \orange{Tan\theta +cot\theta}$

ᴄᴀsᴇ 2:

$\normalsize\star\:\sf \orange{Tan\theta + cos\theta} = \green{sec\theta. cosec\theta}$

$\underline{\bigstar\:\textsf{According \: to \: question:-}}$

sᴏʟᴠɪɴɢ ᴏғ ᴄᴀsᴇ 1:

$\normalsize\hookrightarrow\sf\ sin^2\theta Tan\theta + cos^2\theta cot\theta + 2sin\theta cos\theta = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf\ sin^2\theta Tan\theta + cos^2\theta cot\theta + sin\theta cos\theta + sin\theta cos\theta = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf\ sin^2\theta Tan\theta + sin\theta cos\theta + cos^2\theta cot\theta + sin\theta cos\theta = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf\ sin\theta(sin\theta Tan\theta + cos\theta) + cos\theta(cos\theta cot\theta + sin\theta) = Tan\theta + cot\theta$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ Tan\theta = \frac{sin\theta}{cos\theta} \: and \: cot\theta = \frac{cos\theta}{sin\theta} }) }$

$\normalsize\hookrightarrow\sf\ sin\theta[sin\theta \times\ \frac{sin\theta}{cos\theta} + cos\theta] + cos\theta[cos\theta \times\ \frac{cos\theta}{sin\theta} + sin\theta] = Tan\theta + cot \theta \\ \\ \normalsize\hookrightarrow\sf\ sin\theta[\frac{sin^2\theta}{cos\theta} + cos\theta] + cos\theta[\frac{cos^2\theta}{sin\theta} + sin\theta] = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf\ sin\theta[\frac{sin^2\theta + cos^2\theta}{cos\theta}] + cos[\frac{cos^2\theta + sin^2\theta}{sin\theta}] = Tan\theta + sec\theta$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ sin^2\theta + cos^2\theta = 1}) }$

$\normalsize\hookrightarrow\sf\ sin\theta[\frac{1}{cos\theta}] + cos\theta [\frac{1}{sin\theta}] = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf\ Tan\theta + cot\theta = Tan\theta + cot\theta \\ \\ \normalsize\hookrightarrow\sf \purple{L.H.S \: = \: R.H.S}$

sᴏʟᴠɪɴɢ ᴏғ ᴄᴀsᴇ 2 :

$\normalsize\hookrightarrow\sf\ Tan\theta + cot\theta = sec\theta cosec\theta$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ Tan\theta = \frac{sin\theta}{cos\theta} \: and \: cot\theta = \frac{cos\theta}{sin\theta} }) }$

$\normalsize\hookrightarrow\sf\frac{sin\theta}{cos\theta} + \frac{cos\theta}{sin\theta} = sec\theta cosec\theta \\ \\ \normalsize\hookrightarrow\sf\frac{sin^2\theta + cos^2\theta}{cos\theta sin\theta} = sec\theta cosec\theta$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{ sin^2\theta + cos^2\theta = 1}) }$

$\normalsize\hookrightarrow\sf\frac{1}{sin\theta cos\theta} = sec\theta cosec\theta$

$\scriptsize\sf{\: \: \: \: \: \:( \therefore\ \: \pink{sec\theta = \frac{1}{cos\theta} \: and \: cosec\theta = \frac{1}{cos\theta}}) }$

$\normalsize\hookrightarrow\sf\ sec\theta cosec\theta = sec\theta cosec\theta$

$\normalsize\hookrightarrow\sf \purple{L.H.S \: = \: R.H.S}$