Question

*Trigonometry Class 10*

RD SHARMA (12th Revised 2019)

Exe.11.1

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Q.83. Given that:

(1 + cosA)(1 + cosB)(1 + cosC) = (1 - cosA)(1 - cosB)(1 - cosC)

Show that one of the values of each member of this equality is SinA.SinB.SinC
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**Please explain the meaning of the question first and then answer it.**

Thanks in advance!!

Answer 1

Answer:

(1+cosA)(1+cosB)(1+cosC) = 1 [ dividing both sides with (1-cosA)(1-cosB)(1-cosC)]

(1-cosA)(1-cosB)(1-cosC)

multiplying the numerator and the denominator by (1-cosA)(1-cosB)(1-cosC); we get :

(1+cosA)(1-cosA)(1+cosB)(1-cosB)(1+cosC)(1-cosC) = 1 [ (1-cosA)(1+cosA) = sin2A { sin square A } ]

(1-cosA)(1-cosA)(1-cosB)(1-cosC)(1-cosC)

(sinA.sinB.sinC) = (1-cosA)(1-cosB)(1-cosC)

{ applying square root}

which is obviously equal to (1+cosA)(1+cosB)(1+cosC)

Answer 2

Answer:

Proved!!

Step-by-step explanation:

Given Problem:

Given that:

(1 + cosA)(1 + cosB)(1 + cosC) = (1 - cosA)(1 - cosB)(1 - cosC)

Show that one of the values of each member of this equality is SinA.SinB.SinC

Solution:

To Show:

That one of the values of each member of this equality is SinA.SinB.SinC

Method:

Given that,

$= > ( 1+ cosA) (1 + cosB) (1 + cosC) = (1 - cosA) (1 - cosB) (1 - cosC)$

$= >\dfrac{(1+cosA) (1+cosB)(1+cosC)}{1-cosA(1-cosB)(1-cosC)} = 1$

Now,

Multiply numerator & denominator by, $(1-cosA)(1-cosB)(1-cosC)$

After multiplication we will get,

$= > \dfrac{[(1+cosA)(1+cosB)(1+cosC)]^2}{[(1-cosA)(1-cosB)(1-cosC][(1+cosA)(1+cosB(1+cosC)]} = 1$

$=>\dfrac{[(1+cosA)(1+cosB)(1+cosC)]^2}{sin^2Asin^2Bsin^2C} = 1$

$=>[(1+cosA)(1+cosB)(1+cosC)]^2 = sin^2Asin^2Bsin^2C$

$=>(1+cosA)(1+cosB)(1+cosC) = SinA.SinB.SinC$

Hence Proved!!

Thanks :)