Question

Given that: (1 + cosα) (1 + Cosβ) (1 + cosγ) = (1 - cosα)(1 – cosβ) (1 – cos γ) .Show that one of the values of each member of this equality is sinα sinβ sinγ.

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Answer 1

Given : (1 + cosα) (1 + Cosβ) (1 + cosγ) = (1 - cosα)(1 – cosβ) (1 – cos γ)

To Show : Show that one of the values of each member of this equality is sinα sinβ sinγ.

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$\sf{\bf{Solution \:of\:Question \::}}$

⠀⠀⠀⠀⠀$\implies {\sf{\star (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma ) = (1 - \cos \alpha) (1-\cos \beta )(1-\cos\gamma )}}$

Here ,

• $\implies {\sf{\star L.H.S \:=\: (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma ) }}$

• $\implies {\sf{\star R.H.S \:=\: (1 - \cos \alpha) (1-\cos \beta )(1-\cos\gamma ) }}$

Now By solving both L.H.S and R.H.S:

⠀⠀⠀⠀⠀$\longmapsto {\sf{ (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma ) = (1 - \cos \alpha) (1-\cos \beta )(1-\cos\gamma )}}$

⠀⠀⠀⠀We know that If we have given with the equality of both side we can multiply both [L.H.S and R.H.S] by anyone [L.H.S or R.H.S] .

⠀⠀⠀⠀Now by multiplying both L.H.S and R.H.S by (1 + cosα) (1 + Cosβ) (1 + cosγ) :

⠀⠀⠀⠀$\longmapsto {\sf{ (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma ) = (1 - \cos \alpha) (1-\cos \beta )(1-\cos\gamma )}}$

By multiplying L.H.S we get ,

⠀⠀⠀⠀⠀$\longmapsto {\sf{ (1 + \cos \alpha)^{2} (1+\cos \beta )^{2}(1+\cos\gamma )^{2} = (1 - \cos \alpha) (1-\cos \beta )(1-\cos\gamma )}}$

⠀⠀⠀⠀⠀$\longmapsto {\sf{ ( (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma )) ^{2} = (1 - \cos \alpha) (1-\cos \beta )(1-\cos\gamma )}}$

By multiplying R.H.S we get ,

⠀⠀⠀⠀⠀$\longmapsto {\sf{ ( (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma ))^{2} = (1 - \cos \alpha)(1+\cos\alpha) (1-\cos \beta )(1+\cos\beta)(1-\cos\gamma )(1+\cos\gamma)}}$

As we know that ,

⠀⠀⠀⠀⠀☆ (a + b) (a - b) = a² - b²

⠀⠀⠀⠀⠀$\longmapsto {\sf{ ( (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma )) ^{2} = (1 - \cos^{2} \alpha) (1-\cos^{2} \beta )(1-\cos^{2}\gamma )}}$

As , We know that ,

$\star1 - cos^{2} \theta = \sin^{2} \theta$

⠀⠀⠀⠀⠀$\longmapsto {\sf{ ( (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma )) ^{2} = \sin^{2}\alpha \times \sin^{2} \beta \times \sin^{2} \gamma }}$

Now , By taking root on both side :

⠀⠀⠀⠀⠀$\longmapsto {\sf{ ((1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma )) ^{2} = \sin^{2}\alpha \times \sin^{2} \beta \times \sin^{2} \gamma }}$

By taking root in L.H.S and R.H.S :

⠀⠀⠀⠀⠀$\longmapsto {\sf{ \sqrt {( (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma )) ^{2} }= \sqrt{ \sin^{2}\alpha \times \sin^{2} \beta \times \sin^{2} \gamma}}}$

As , We know that ,

$\star\sqrt {y^{2}} = y$

⠀⠀⠀⠀⠀$\longmapsto {\sf{ (1 + \cos \alpha) (1+\cos \beta )(1+\cos\gamma )= sin\alpha \times \sin \beta \times \sin \gamma }}$

⠀⠀⠀⠀⠀Now , it is proved that

⠀⠀⠀⠀⠀Each member's value of L.H.S is equal to sinα × sinβ × sinγ

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$\large{\boxed {\mathrm |\:\:{\underline {More\:To\:Know\::}}\:\:|}}$

Trigonometric Identities :

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

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Answer 2

Answer:

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