Question

see the attachment nd plz solve it....

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Answer 1

Answer:

$\color{Red}LHS= \cos {}^{2}x + (x + \frac{\pi}{3}) + \cos {}^{2}(x - \frac{\pi}{3} )$

$\color{Pink} \frac{1}{2}(1 + \cos2x) + \frac{1}{2} [1 + \cos2 (x + \frac{\pi}{3})] + \frac{1}{2}[1 + \cos2(x - \frac{\pi}{3})]$

$\color{Orange} \frac{1}{2}[3 + \cos2x + \cos(2x + \frac{2\pi}{3}) + \cos( 2x - \frac{2\pi}{3}) ]$

$\color{SkyBlue} [3 + \cos2x + 2 \: \cos2x \cos \frac{2\pi}{3}]$

$\color{Purple} \frac{1}{2}[3 + \cos2x + 2( \cos2x) \: \: ( \frac{ - 1}{2}) ]$

$\color{Red} \frac{1}{2}[3 + \cos2x - \cos2x\: \: ] = \frac{3}{2} = RHS$

Answer 2

Question:-

Attachment.

Correct answer:-

Answer:

\color{Red}LHS= \cos {}^{2}x + (x + \frac{\pi}{3}) + \cos {}^{2}(x - \frac{\pi}{3} )LHS=cos

2

x+(x+

3

π

)+cos

2

(x−

3

π

)

\color{Pink} \frac{1}{2}(1 + \cos2x) + \frac{1}{2} [1 + \cos2 (x + \frac{\pi}{3})] + \frac{1}{2}[1 + \cos2(x - \frac{\pi}{3})]

2

1

(1+cos2x)+

2

1

[1+cos2(x+

3

π

)]+

2

1

[1+cos2(x−

3

π

)]

\color{Orange} \frac{1}{2}[3 + \cos2x + \cos(2x + \frac{2\pi}{3}) + \cos( 2x - \frac{2\pi}{3}) ]

2

1

[3+cos2x+cos(2x+

3

2π

)+cos(2x−

3

2π

)]

\color{SkyBlue} [3 + \cos2x + 2 \: \cos2x \cos \frac{2\pi}{3}][3+cos2x+2cos2xcos

3

2π

]

\color{Purple} \frac{1}{2}[3 + \cos2x + 2( \cos2x) \: \: ( \frac{ - 1}{2}) ]

2

1

[3+cos2x+2(cos2x)(

2

−1

)]

\color{Red} \frac{1}{2}[3 + \cos2x - \cos2x\: \: ] = \frac{3}{2} = RHS

2

1

[3+cos2x−cos2x]=

2

3

=RHS

Thnkuuu...!!! :)

hope it helps...!!! ♡

#ItzzMichhAditi ♡