Question

$\pink{\boxed{\bf Question :}}$

m = cosec theta + cot theta show that m square - 1/m square + 1 = cos theta.

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Answer 1

Given :-

• $\sf m = cosec \: \theta + cot \: \theta$

To show :-

• $\sf \dfrac{m^{2} - 1}{m^{2} + 1} = cos \: \theta$

Prove :-

$\sf We \: have :$

• $\sf \dfrac{m^{2} - 1}{m^{2} + 1} = cos \: \theta$

$\underline{\sf Take \: L.H.S :}$

$\implies \sf \dfrac{m^{2} - 1}{m^{2} + 1}$

• $\sf Put \: m = cosec \: \theta + cot \: \theta$

$\implies \sf \dfrac{(cosec \: \theta + cot \: \theta ) ^{2} - 1}{(cosec \: \theta + cot \: \theta)^{2} - 1}$

• $\sf cosec^{2} \: \theta - cot^{2} \: \theta = 1$

$\implies \sf \dfrac{(cosec \: \theta + cot \: \theta)^{2} - (cosec^{2} \: \theta - cot^{2} \: \theta)}{(cosec \: \theta + cot \: \theta)^{2} + (cosec^{2} \: \theta - cot^{2} \: \theta)}$

• Use a² - b² = (a + b)(a - b) in numerator and denominator.

$\implies \sf \dfrac{(cosec \: \theta + cot \: \theta)^{2} - \bigg( (cosec \: \theta + cot \: \theta)( cosec \: \theta - cot \: \theta)\bigg)}{(cosec \: \theta + cot \: \theta)^{2} + \bigg( (cosec \: \theta + cot \: \theta)(cosec \: \theta - cot \: \theta) \bigg)}$

• Take cosec $\theta$ + cot $\theta$ common from numerator and denominator.

$\implies \sf \dfrac{\cancel{(cosec \: \theta + cot \: \theta)} \times \bigg( (cosec \: \theta + cot \: \theta) - ( cosec \: \theta - cot \: \theta) \bigg)}{\cancel{(cosec \: \theta + cot \: \theta)} \times \bigg( (cosec \: \theta + cot \: \theta) + (cosec \: \theta - cot \: \theta)\bigg)}$

$\implies \sf \dfrac{\cancel{cosec \: \theta} + cot \: \theta - \cancel{ cosec \: \theta} + cot \: \theta}{cosec \: \theta + \cancel{ cot \: \theta} + cosec \: \theta - \cancel{ cot \: \theta}}$

$\implies \sf \dfrac{\cancel{2} \: cot \: \theta}{\cancel{2} \: cosec \: \theta}$

• cot $\theta$ = cos $\theta$/sin $\theta$ and cosec $\theta$ = 1/sin $\theta$

$\implies \sf \dfrac{\dfrac{cos \: \theta}{\cancel{sin \: \theta}}}{\dfrac{1}{\cancel{sin \: \theta}}}$

$\implies \sf cos \: \theta$

$\implies \sf R.H.S$

Hence, Proved !!

Answer 2

Answer:

Given :-

• \sf m = cosec \: \theta + cot \: \thetam=cosecθ+cotθ

To show :-

• \sf \dfrac{m^{2} - 1}{m^{2} + 1} = cos \: \theta

m

2

+1

m

2

−1

=cosθ

Prove :-

\sf We \: have :Wehave:

• \begin{gathered} \sf \dfrac{m^{2} - 1}{m^{2} + 1} = cos \: \theta \\ \\ \end{gathered}

m

2

+1

m

2

−1

=cosθ

\underline{\sf Take \: L.H.S :}

TakeL.H.S:

\begin{gathered} \implies \sf \dfrac{m^{2} - 1}{m^{2} + 1} \\ \\ \end{gathered}

⟹

m

2

+1

m

2

−1

\sf Put \: m = cosec \: \theta + cot \: \thetaPutm=cosecθ+cotθ

\begin{gathered} \implies \sf \dfrac{(cosec \: \theta + cot \: \theta ) ^{2} - 1}{(cosec \: \theta + cot \: \theta)^{2} - 1} \\ \\ \end{gathered}

⟹

(cosecθ+cotθ)

2

−1

(cosecθ+cotθ)

2

−1

\sf cosec^{2} \: \theta - cot^{2} \: \theta = 1cosec

2

θ−cot

2

θ=1

\begin{gathered} \implies \sf \dfrac{(cosec \: \theta + cot \: \theta)^{2} - (cosec^{2} \: \theta - cot^{2} \: \theta)}{(cosec \: \theta + cot \: \theta)^{2} + (cosec^{2} \: \theta - cot^{2} \: \theta)} \\ \\ \end{gathered}

⟹

(cosecθ+cotθ)

2

+(cosec

2

θ−cot

2

θ)

(cosecθ+cotθ)

2

−(cosec

2

θ−cot

2

θ)

Use a² - b² = (a + b)(a - b) in numerator and denominator.

\begin{gathered} \implies \sf \dfrac{(cosec \: \theta + cot \: \theta)^{2} - \bigg( (cosec \: \theta + cot \: \theta)( cosec \: \theta - cot \: \theta)\bigg)}{(cosec \: \theta + cot \: \theta)^{2} + \bigg( (cosec \: \theta + cot \: \theta)(cosec \: \theta - cot \: \theta) \bigg)} \\ \\ \end{gathered}

⟹

(cosecθ+cotθ)

2

+((cosecθ+cotθ)(cosecθ−cotθ))

(cosecθ+cotθ)

2

−((cosecθ+cotθ)(cosecθ−cotθ))

Take cosec \thetaθ + cot \thetaθ common from numerator and denominator.

\begin{gathered} \implies \sf \dfrac{\cancel{(cosec \: \theta + cot \: \theta)} \times \bigg( (cosec \: \theta + cot \: \theta) - ( cosec \: \theta - cot \: \theta) \bigg)}{\cancel{(cosec \: \theta + cot \: \theta)} \times \bigg( (cosec \: \theta + cot \: \theta) + (cosec \: \theta - cot \: \theta)\bigg)} \\ \\ \end{gathered}

⟹

(cosecθ+cotθ)

×((cosecθ+cotθ)+(cosecθ−cotθ))

(cosecθ+cotθ)

×((cosecθ+cotθ)−(cosecθ−cotθ))

\begin{gathered} \implies \sf \dfrac{\cancel{cosec \: \theta} + cot \: \theta - \cancel{ cosec \: \theta} + cot \: \theta}{cosec \: \theta + \cancel{ cot \: \theta} + cosec \: \theta - \cancel{ cot \: \theta}} \\ \\ \end{gathered}

⟹

cosecθ+

cotθ

+cosecθ−

cotθ

cosecθ

+cotθ−

cosecθ

+cotθ

\begin{gathered} \implies \sf \dfrac{\cancel{2} \: cot \: \theta}{\cancel{2} \: cosec \: \theta} \\ \\ \end{gathered}

⟹

2

cosecθ

2

cotθ

cot \thetaθ = cos \thetaθ /sin \thetaθ and cosec \thetaθ = 1/sin \thetaθ

\begin{gathered} \implies \sf \dfrac{\dfrac{cos \: \theta}{\cancel{sin \: \theta}}}{\dfrac{1}{\cancel{sin \: \theta}}} \\ \\ \end{gathered}

⟹

sinθ

1

sinθ

cosθ

\begin{gathered} \implies \sf cos \: \theta \\ \\ \end{gathered}

⟹cosθ

\begin{gathered} \implies \sf R.H.S \\ \\ \end{gathered}

⟹R.H.S

Hence, Proved !!