Prove that-i cos-cos-cos-cos -4 cos-2 -cos-y-2 -cosy-2-ii If tan x - b - a- then -a- b - a - b - a - b -a- b -2 cos x -cos 2 x -

LHS = cos α+cos β+cos γ+cos (α+β+γ) = (cos α+cos β)+[cos γ+cos (α+β+γ) ] = 2cos(α+β/2 ).cos(α β/2 )+ 2cos(γ+α+β+γ/2 ). cos(γ α β γ/2 ) [∵ cos x+cos y=2cos(x+y/ ...

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Byju's Answer

Standard XI

Mathematics

Conditional Identities

Prove that,i ...

Question

Prove that,

(i) $c o s α + c o s β + c o s γ + c o s (α + β + γ) = 4 c o s (\frac{α + β}{2}) . c o s (\frac{β + γ}{2}) . c o s (\frac{γ + α}{2}) .$

(ii) If $t a n x = \frac{b}{a},$ then $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = \frac{2 c o s x}{\sqrt{c o s 2 x}} .$

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Solution

LHS = $c o s α + c o s β + c o s γ + c o s (α + β + γ)$

= $(c o s α + c o s β) + [c o s γ + c o s (α + β + γ)]$

= $2 c o s (\frac{α + β}{2}) . c o s (\frac{α - β}{2}) + 2 c o s (\frac{γ + α + β + γ}{2}) . c o s (\frac{γ - α - β - γ}{2})$

$[∵ c o s x + c o s y = 2 c o s (\frac{x + y}{2}) c o s (\frac{x - y}{2})]$

= $2 c o s (\frac{α + β}{2}) . c o s (\frac{α - β}{2}) + 2 c o s (\frac{α + β + 2 γ}{2}) . c o s (\frac{α + β}{2})$

$[∵ c o s (- x) = c o s x]$

= $2 c o s (\frac{α + β}{2}) [c o s (\frac{α - β}{2}) + c o s (\frac{α + β + 2 γ}{2})]$

= $2 c o s (\frac{α + β}{2}) [2 c o s (\frac{\frac{α - β}{2} + \frac{α + β + 2 γ}{2}}{2}) . c o s (\frac{\frac{α - β}{2} - \frac{α + β + 2 γ}{2}}{2})]$

= $2 c o s (\frac{α + β}{2}) [2 c o s (\frac{α + γ}{2}) . c o s (\frac{- β - γ}{2})]$

= $2 c o s (\frac{α + β}{2}) [2 c o s (\frac{α + γ}{2}) . c o s {(- \frac{(β + γ)}{2})}]$

= $2 c o s (\frac{α + β}{2}) [2 c o s (\frac{α + γ}{2}) . c o s (\frac{β + γ}{2})]$

$[∵ c o s (- x) = c o s x]$

= $4 c o s (\frac{α + β}{2}) . c o s (\frac{β + γ}{2}) . c o s (\frac{α + γ}{2})$

= RHS Hence proved.

(ii) $L H S = \sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}$

[ $∵$ dividing by numerator and denominator by a]

= $\frac{1 + \frac{b}{a} + 1 - \frac{b}{a}}{\sqrt{(1 - \frac{b}{a}) (1 + \frac{b}{a})}}$

= $\frac{2}{\sqrt{1 - \frac{b^{2}}{a^{2}}}} = \frac{2}{\sqrt{1 - t a n^{2} x}}$ $[∵ t a n x = \frac{b}{a}]$

= $\frac{2}{\sqrt{1 - \frac{s i n^{2} x}{c o s^{2} x}}} = \frac{2 c o s x}{\sqrt{c o s^{2} x - s i n^{2} x}}$

= $\frac{2 c o s x}{\sqrt{c o s 2 x}} = R H S [∵ c o s^{2} θ - s i n^{2} θ = c o s 2 θ]$

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Similar questions

Q. If

a = cos α + i sin α, b = cos β + i sin β,

c = cos γ + i sin γ

and

\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1,

then

Q. Show that

cos α + cos β + cos γ = cos (α + β + γ) = 4 cos {(α + β) / 2} cos {(β + γ) / 2} cos {(γ + α) / 2}

Q. If

a = cos α + i sin α, b = cos β + i sin β, c = cos γ + i sin γ

and

\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = - 1

, then

cos (β - γ) + cos (γ - α) + cos (α - β) =

cos α + cos β + cos γ + cos (α + β + γ) = b cos \frac{α + β}{2} cos \frac{β + γ}{2} cos \frac{γ + α}{2}

. Find

b

cos (α - β) + cos (β - γ) + cos (γ - α) = - \frac{3}{2}

then prove that

cos α + cos β + cos γ = sin α + sin β + sin γ = 0

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Prove that, (i) cos α+cos β+cos γ+cos(α+β+γ)=4 cos(α+β2). cos(β+γ2). cos(γ+α2). (ii) If tan x=ba, then √a+ba−b+√a−ba+b=2 cos x√cos 2x.

Prove that,

(i) $c o s α + c o s β + c o s γ + c o s (α + β + γ) = 4 c o s (\frac{α + β}{2}) . c o s (\frac{β + γ}{2}) . c o s (\frac{γ + α}{2}) .$

(ii) If $t a n x = \frac{b}{a},$ then $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = \frac{2 c o s x}{\sqrt{c o s 2 x}} .$