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VirαtKσhli

Correct Question :-

If sec[tex]\theta[/tex] + tan[tex]\theta[/tex]= x , then prove that ,

[tex]\implies\sf sin\theta =\dfrac{x^2-1}{x^2+1}[/tex]

Proof :-

Here we are given that ,

[tex]\longrightarrow \sec\theta + tan\theta = x[/tex]

Firstly write everything in terms of sine and cosine .

[tex]\longrightarrow \dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}=x [/tex]

Add ,

[tex]\longrightarrow \dfrac{1+\sin\theta}{\cos\theta}=x [/tex]

On squaring both sides , we have ;

[tex]\longrightarrow \dfrac{(\sin\theta+1)^2}{(\cos\theta)^2}=x^2[/tex]

Simplify using identity sin²x + cos²x = 1 ,

[tex]\longrightarrow \dfrac{(1+\sin\theta)^2}{1-\sin^2\theta}=x^2 [/tex]

Simplify using identity (a+b)(a-b)=a²-b² ,

[tex]\longrightarrow \dfrac{(1+\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)}=x^2 [/tex]

Simplify,

[tex]\longrightarrow \dfrac{1+\sin\theta}{1-\sin\theta}=x^2 [/tex]

On using Componendo and Dividendo , we have ;

[tex]\longrightarrow \dfrac{1+\sin\theta+1-\sin\theta}{1+\sin\theta-1+\sin\theta}=\dfrac{x^2+1}{x^2-1}[/tex]

[tex]\longrightarrow \dfrac{2}{2\sin\theta}=\dfrac{x^2+1}{x^2-1}[/tex]

Simplify,

[tex]\longrightarrow \dfrac{1}{\sin\theta}=\dfrac{x^2+1}{x^2-1}\\[/tex]

Divide both the sides by 1 ,

[tex]\longrightarrow \underline{\underline{\sin\theta =\dfrac{x^2-1}{x^2+1}}} [/tex]

Hence proved .

And we are done !

[tex]\rule{200}4[/tex]

More to Know :-

1) Trigonometric table :-

[tex]\small{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}[/tex]

[tex]\rule{200}4[/tex]

2) Important identities :-

[tex]\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ $\: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1$\end{minipage}}[/tex]

[tex]\rule{200}4[/tex]

Question:

If sec theta + tan theta equals to x then prove that:-

[tex]\longrightarrow \: sin \theta = \frac{ {x}^{2} - { \bold{1}} }{ {x}^{2} + 1 } [/tex]

Solution:

Given that:

[tex]\longrightarrow \: sec \: \theta + tan \: \theta =x \: \: ..(i) [/tex]

We have to prove:

[tex]\longrightarrow \: sin \: \theta = \frac{ {x}^{2} - 1 }{ {x}^{2} + 1} [/tex]

We know that:

[tex]\longrightarrow \: { \sec}^{2} \theta \: - \: { \tan}^{2} \theta = 1[/tex]

[tex] \longrightarrow ( \sec \: \theta \: + \: \tan \: \theta)( \sec \: \theta - \tan \: \theta )[/tex]

[tex] \longrightarrow x( \sec \: \theta - \tan \: \theta) \: = 1[/tex]

[tex] \longrightarrow \: \sec \theta - \tan \: \theta = \frac{1}{x} \ ..(ii)[/tex]

Adding equation (i) and (ii), we get:

[tex] \longrightarrow \: 2\sec \theta = x + \frac{1}{x} [/tex]

[tex] \longrightarrow \: 2 \sec \: \theta = \frac{ {x}^{2} + 1}{x} [/tex]

[tex] \longrightarrow \: \sec \: \theta = \frac{ {x}^{2} + 1}{2x} [/tex]

[tex] \longrightarrow \: \cos \: \theta = \frac{2x}{ {x}^{2} + 1 } [/tex]

Now, we know that:

[tex] \longrightarrow \: { \sin}^{2} \theta + { \cos }^{2} \: \theta = 1[/tex]

Therefore,

[tex] \longrightarrow \: \sin \theta \: = \sqrt{1 - { \cos }^{2} } \: \theta[/tex]

[tex] \longrightarrow \: \sin \: \theta = {\sqrt{1 - ( \frac{2x}{ {x}^{2} + 1} } )}^{2} [/tex]

[tex] \longrightarrow \: \sin \: \theta = \sqrt{ \frac{ {({x}^{2} + 1)}^{2} - {(2x)}^{2} }{( {x}^{2} + 1)} } [/tex]

[tex] \longrightarrow \: \sin \: \theta = \sqrt{ \frac{ {x}^{4} \: + 2 {x}^{2} + 1 - \: 4 {x}^{2} }{( {x}^{2} { + 1)}^{2} } } [/tex]

[tex] \longrightarrow \: \sin \: \theta = \sqrt{ \frac{ {x}^{4} - {2x}^{2} + 1 }{( {x}^{2} { + 1)}^{2} } } [/tex]

[tex] \longrightarrow \: \sin \: \theta = \sqrt{ \frac{( {x}^{2} { - 1)}^{2} }{ {(x}^{2} + {1)}^{2} } } [/tex]

[tex] \longrightarrow \: \sin \: \theta = \frac{ {x}^{2} - 1 }{ {x}^{2} + 1} [/tex]

Hence Proved..!!

[tex] \: [/tex]

Learn More:

1. Relationship between sides and T-Ratios.

  • sin θ = Height/Hypotenuse
  • cos θ = Base/Hypotenuse
  • tan θ = Height/Base
  • cot θ = Base/Height
  • sec θ = Hypotenuse/Base
  • cosec θ = Hypotenuse/Height

2. Square formulae.

  • sin²θ + cos²θ = 1
  • cosec²θ - cot²θ = 1
  • sec²θ - tan²θ = 1

3. Reciprocal Relationship.

  • sin θ = 1/cosec θ
  • cos θ = 1/sec θ
  • tan θ = 1/cot θ
  • cosec θ = 1/sin θ
  • sec θ = 1/cos θ
  • tan θ = 1/cot θ

4. Cofunction identities.

  • sin(90° - θ) = cos θ
  • cos(90° - θ) = sin θ
  • cosec(90° - θ) = sec θ
  • sec(90° - θ) = cosec θ
  • tan(90° - θ) = cot θ
  • cot(90° - θ) = tan θ

5. Even odd identities.

  • sin -θ = -sin θ
  • cos -θ = cos θ
  • tan -θ = -tan θ

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