Question

can you solve step by step
don't spam​

Answer 1

Correction in the question:-

$\rm tan \theta + \underline{cot \theta} = sec \theta \cdot cosec \theta$

Solution:-

$\rm tan \theta +cot \theta= sec \theta \cdot cosec \theta$

Since,

$\tan \theta = \dfrac{\sin \theta}{\cos \theta} \ and \ \cot \theta= \dfrac{\cos \theta}{\sin \theta}, \ \therefore$

$\rm\dashrightarrow \dfrac{sin\theta}{cos\theta} + \dfrac{cos\theta}{sin\theta}= sec \theta \cdot cosec \theta$

$\rm\dashrightarrow \dfrac{(sin\theta)(sin\theta)+(cos\theta)(cos\theta)}{(cos\theta)(sin\theta)}= sec \theta \cdot cosec \theta$

$\rm\dashrightarrow \dfrac{sin^2\theta+cos^2\theta}{(cos\theta)(sin\theta)}= sec \theta \cdot cosec \theta$

And we know that,

sin² θ + cos² θ = 1

Hence,

$\rm\dashrightarrow \dfrac{1}{(cos\theta)(sin\theta)}= sec \theta \cdot cosec \theta$

$\rm\dashrightarrow \dfrac{1}{cos\theta}\cdot\dfrac{1}{sin\theta}= sec \theta \cdot cosec \theta$

We know that

$\bullet\quad\dfrac{1}{\cos \theta} = \sec \theta$

$\bullet\quad\rm\dfrac{1}{\sin \theta} = cosec \ \theta$

∴,

$\bf\dashrightarrow sec\theta\cdot cosec\theta= sec \theta \cdot cosec \theta$

Hence, Proved.

Answer 2

Answer:

$\begin{gathered}\large \mathbb \blue{TO \: PROVE : - } \\\end{gathered}TOPROVE:−</p><p></p><p>\begin{gathered}\sf \: \frac{1}{sec \theta + tan \theta} = {sec \theta + tan \theta} \\\end{gathered}secθ+tanθ1=secθ+tanθ</p><p></p><p>\begin{gathered}\large \mathbb \blue{SOLUTION: - } \\\end{gathered}SOLUTION:−</p><p></p><p>\begin{gathered}\sf \: \: L.H.S. \: \: = \frac{1}{sec \theta + tan \theta} \\\end{gathered}L.H.S.=secθ+tanθ1</p><p></p><p>\begin{gathered}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{{sec \theta + tan \theta}}{({sec \theta - tan \theta}) ({sec \theta + tan \theta}) } \\\end{gathered}=(secθ−tanθ)(secθ+tanθ)secθ+tanθ</p><p></p><p>\begin{gathered}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: = \frac{sec \theta + tan \theta}{ {sec}^{2} \theta - {tan}^{2} \theta } \\\end{gathered}=sec2θ−tan2θsecθ+tanθ</p><p></p><p>\begin{gathered}\star \: \rm \red{\: by \: algebric \: identity} \\\end{gathered}⋆byalgebricidentity</p><p></p><p>\begin{gathered}\green{ \boxed{ \sf \: (a + b)(a - b) = {a}^{2} - {b}^{2} }} \\\end{gathered}(a+b)(a−b)=a2−b2</p><p></p><p>\begin{gathered}\longrightarrow \sf \: \frac{sec \theta + tan \theta}{1} \\\end{gathered}⟶1secθ+tanθ</p><p></p><p>\begin{gathered}\star \: \rm \red{\: by \: trignometric \: identity} \\\end{gathered}⋆bytrignometricidentity</p><p></p><p>\begin{gathered}\sf \green{\boxed { \sf {sec}^{2} A - {tan}^{2} A = 1}} \\\end{gathered}sec2A−tan2A=1</p><p></p><p>\begin{gathered}\implies \sf \: sec \theta + tan \theta = R.H.S. \\\end{gathered}⟹secθ+tanθ=R.H.S.</p><p></p><p>\begin{gathered}\tt \: therefore. \\\end{gathered}therefore.</p><p></p><p>\begin{gathered}\boxed{\sf \: \frac{1}{sec \theta + tan \theta} = {sec \theta + tan \theta} }\\\end{gathered}secθ+tanθ1=secθ+tanθ</p><p></p><p>\begin{gathered}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frak \purple{ \: hence \: proved} \: \: \: \: \: \: \: \\\end{gathered}henceproved</p><p></p><p>$