Question

Show that cos 2 (45 + theta) + cos 2 (45 - theta) / tan (60 + theta) tan (30 - theta) = 1

Answer 1

Answer:-

Given:-

$\frac{cos^2(45\°+\theta)+cos^2(45\°-\theta)}{tan(60\°+\theta).tan(30\°-\theta)}$

To Prove:-

$\frac{cos^2(45\°+\theta)+cos^2(45\°-\theta)}{tan(60\°+\theta).tan(30\°-\theta)} =1$

Solution:-

we will take,

⇒ $L.H.S = \frac{cos^2(45\°+\theta)+cos^2(45\°-\theta)}{tan(60\°+\theta).tan(30\°-\theta)}$

⇒ $\frac{cos^2(45\°+\theta)[sin(90\°-(45\°-\theta))]}{tan(60\°+\theta).cot(90\°-(30\°-\theta))}$

                      $[sin(90\°-\theta)=cos\theta \: and \: cot(90\°-\theta)=tan\theta]$

⇒ $\frac{cos^2(45\°+\theta)+sin^2(45\°+\theta)}{tan(60\°+\theta).cot(60\°+\theta)}$

                      $[sin^2\theta+cos^2\theta=1]$

⇒ $\frac{1}{tan(60\°+\theta).\frac{1}{tan(60\°+\theta)} } = 1.$

                      $[cot\theta=\frac{1}{tan\theta} ]$

⇒ $L.H.S=R.H.S$

$Hence, \: Proved$

Answer 2

Answer

LHS =cos^2(45 + theta) + sin^2{90 - (45 -theta)} / tan(60 + theta) cot{90 - (30 - theta)}

=cos^2(45 + theta) + sin^2(90 - 45 + theta) / tan(60 + theta) cot(90 - 30 + theta)

=cos^2(45 + theta) + sin^2(45 + theta) / tan(60 + theta) cot(60 + theta)

= 1 / 1

LHS = 1 = RHS