Question

(2) . 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

Step-by-step explanation:

i) The curved surface area of cylinderical petrol tank is 59.4 m².

ii) The steel actually used to make the tank is 95.04 m².

Step-by-step-explanation:

We have given that,

For a cylinderical petrol tank,

Diameter ( d ) = 4.2 m

∴ Radius ( r ) = d ÷ 2 = 4.2 ÷ 2 = 2.1 m

Height ( h ) = 4.5 m

We have to find,

i) Curved surface area of tank

ii) Total steel used to make tank

i)

Now, we know that,

\displaystyle{\pink{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:h}}Curvedsurfaceareaofcylinder=2πrh

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:\times\:4.5}⟹CSAcylinder=2×722×2.1×4.5

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:4.5}⟹CSAcylinder=2×22×0.3×4.5

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:9}⟹CSAcylinder=22×0.3×9

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:198\:\times\:0.3}⟹CSAcylinder=198×0.3

\displaystyle{\implies\sf\:CSA_{cylinder}\:=\:59.4\:m^2}⟹CSAcylinder=59.4m2

\therefore\:\underline{\boxed{\red{\sf\:Curved\:surface\:area\:of\:cylinder\:=\:59.4\:m^2\:}}}∴Curvedsurfaceareaofcylinder=59.4m2

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ii)

Now,

The tank is closed with steel.

We have to find how much steel was used to make the tank.

Now,

\displaystyle{\pink{\sf\:Total\:surface\:area\:of\:cylinder\:=\:2\:\pi\:r\:(\:r\:+\:h\:)}}Totalsurfaceareaofcylinder=2πr(r+h)

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:\dfrac{22}{\cancel{7}}\:\times\:\cancel{2.1}\:(\:2.1\:+\:4.5\:)}⟹TSAcylinder=2×722×2.1(2.1+4.5)

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:2\:\times\:22\:\times\:0.3\:\times\:6.6}⟹TSAcylinder=2×22×0.3×6.6

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:0.3\:\times\:13.2}⟹TSAcylinder=22×0.3×13.2

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:22\:\times\:3.96}⟹TSAcylinder=22×3.96

\displaystyle{\implies\sf\:TSA_{cylinder}\:=\:87.12\:m^2}⟹TSAcylinder=87.12m2

\therefore\:\underline{\boxed{\green{\sf\:Total\:surface\:area\:of\:cylinder\:=\:87.12\:m^2\:}}}∴Totalsurfaceareaofcylinder=87.12m2

Now, from the given condition,

\displaystyle{\pink{\sf\:Steel\:used\:to\:make\:tank\:-\:Steel\:wasted\:=\:Total\:surface\:area\:of\:cylinder}}Steelusedtomaketank−Steelwasted=Totalsurfaceareaofcylinder

\displaystyle{\implies\sf\:Steel\:used\:-\:\dfrac{1}{12}\:Steel\:used\:=\:TSA_{cylinder}}⟹Steelused−121Steelused=TSAcylinder

\displaystyle{\implies\sf\:Steel\:used\:\left(\:1\:-\:\dfrac{1}{12}\:\right)\:=\:87.12}⟹Steelused(1−121)=87.12

\displaystyle{\implies\sf\:Steel\:used\:\left(\:\dfrac{12\:-\:1}{12}\:\right)\:=\:87.12}⟹Steelused(1212−1)=87.12

\displaystyle{\implies\sf\:Steel\:used\:\times\:\dfrac{11}{12}\:=\:87.12}⟹Steelused×1211=87.12

\displaystyle{\implies\sf\:Steel\:used\:=\:\dfrac{\cancel{87.12}\:\times\:12}{\cancel{11}}}⟹Steelused=

Aladdin naam toh suna hoga...

Answer 2

★ 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.

★ Solution:

We have,

➡ (1 + cosα)(1 + Cosβ)(1 + cosγ) = (1 - cosα)(1 – cosβ)(1 – cosγ)

Multiplying both sides by (1 + cosα)(1 + Cosβ)(1 + cosγ), we get,

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)]² = (1 - cosα)(1 – cosβ)(1 – cosγ) × (1 + cosα)(1 + Cosβ)(1 + cosγ)

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)]² = (1 - cos²α)(1 – cos²β)(1 – cos²γ)

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)]² = (sin²α)(sin²β)(sin²γ)

➡ [(1 + cosα)(1 + Cosβ)(1 + cosγ)] = ± sinα sinβ sinγ

➡ Hence, one of the values of (1 + cosα) (1 + Cosβ) (1 + cosγ) is sinα sinβ sinγ

Again, we have,

➡ (1 + cosα)(1 + Cosβ)(1 + cosγ) = (1 - cosα)(1 – cosβ)(1 – cosγ)

Multiplying both sides by (1 - cosα)(1 – cosβ)(1 – cosγ), we get,

➡ (1 + cosα)(1 + Cosβ)(1 + cosγ)(1 - cosα)(1 – cosβ)(1 – cosγ) = [(1 - cosα)(1 – cosβ)(1 – cosγ)]²

➡ [(1 - cosα)(1 – cosβ)(1 – cosγ)]² = (1 - cos²α)(1 – cos²β)(1 – cos²γ)

➡ [(1 - cosα)(1 – cosβ)(1 – cosγ)]² = (sin²α)(sin²β)(sin²γ)

➡ (1 - cosα)(1 – cosβ)(1 – cosγ) = ± (sinα)(sinβ)(sinγ)

➡ Hence, one of the values of (1 - cosα)(1 – cosβ)(1 – cosγ) is (sinα)(sinβ)(sinγ)

Thus, one of the values of each member of this equality is (sinα)(sinβ)(sinγ).

(Hence Proved)

Formula Used:

sin²(x) = 1 - cos²(x)

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