Question

see the attachment nd plz solve it....​

nd don't spam otherwise your answer will be reported..​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cos^2(x)+cos^2\bigg(x+\dfrac{\pi}{3}\bigg)+cos^2\bigg(x-\dfrac{\pi}{3}\bigg)}$

$\sf{=cos^2(x)+\bigg\{cos(x)\,cos\bigg(\dfrac{\pi}{3}\bigg)-sin(x)\,sin\bigg(\dfrac{\pi}{3}\bigg)\bigg\}^2+\bigg\{cos(x)\,cos\bigg(\dfrac{\pi}{3}\bigg)+sin(x)\,sin\bigg(\dfrac{\pi}{3}\bigg)\bigg\}^2}$

$\sf{=cos^2(x)+\bigg\{\dfrac{1}{2}\cdot\,cos(x)-\dfrac{\sqrt{3} }{2}\cdot\,sin(x)\bigg\}^2+\bigg\{\dfrac{1}{2}\cdot\,cos(x)+\dfrac{\sqrt{3} }{2}\cdot\,sin(x)\bigg\}^2}$

$\rm{\green{We\,\,know,\,\,(a+b)^2+(a-b)^2=2(a^2+b^2)}}$

So,

$\sf{=cos^2(x)+2\bigg[\bigg\{\dfrac{1}{2}\cdot\,cos(x)\bigg\}^2+\bigg\{\dfrac{\sqrt{3} }{2}\cdot\,sin(x)\bigg\}^2\bigg]}$

$\sf{=cos^2(x)+2\bigg[\dfrac{1}{4}\cdot\,cos^2(x)+\dfrac{3}{4}\cdot\,sin^2(x)\bigg]}$

$\sf{=cos^2(x)+\dfrac{1}{2}\cdot\,cos^2(x)+\dfrac{3}{2}\cdot\,sin^2(x)}$

$\sf{=\bigg(1+\dfrac{1}{2}\bigg)cos^2(x)+\dfrac{3}{2}sin^2(x)}$

$\sf{=\dfrac{3}{2}cos^2(x)+\dfrac{3}{2}sin^2(x)}$

$\sf{=\dfrac{3}{2}\{cos^2(x)+sin^2(x)\}}$

$\sf{=\dfrac{3}{2}\times1}$

$\sf{=\dfrac{3}{2}}$

Answer 2

Answer:

me aap ko achi trhse janti nihu is lia