Question

show that cos 2(45°+θ)+cos2(45°−θ)/tan(60°+θ)tan(30°−θ)= 1​

Answer 1

Answer:

Step-by-step explanation:

LHS = cos ²(45°+ϴ) + cos²(45° - ϴ) / tan(60° +ϴ) tan(30 - ϴ) = 1

= sin²[90° - (45°+ϴ)]+ cos²(45° - ϴ) / cot [90° - (60° +ϴ)] tan(30 - ϴ)

[ Cosϴ= sin(90°-ϴ) & cot ϴ= tan (90°-ϴ)]

= sin²(45°-ϴ)+ cos²(45° - ϴ) / cot 30° - ϴ)] tan(30° - ϴ)

= 1/1 = RHS

[ sin²ϴ + cos²ϴ= 1 and tanϴcotϴ=1]

HOPE THIS WILL HELP YOU...

Answer 2

Solution:

To Show:

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

tan(60+θ).tan(30−θ)

cos

2

(45+θ)+cos

2

(45−θ)

=1

Formulas Used:

\begin{gathered}cosA=sin(90-A)\\\\tanA=cot(90-A)\\\\cos^2A+sin^2A=1\\\\tanA.cotA=1\end{gathered}

cosA=sin(90−A)

tanA=cot(90−A)

cos

2

A+sin

2

A=1

tanA.cotA=1

Proof:

Taking Left Hand Side,\begin{gathered}\dfrac{cos^2(45+\theta)+cos^2(45-\theta)}{tan(60+\theta).tan(30-\theta)}\\\\\\=\dfrac{cos^(45+\theta)+sin^2(90-45+\theta)}{tan(60+\theta).cot(90-30+\theta)}\\\\\\=\dfrac{cos^2(45+\theta)+sin^2(45+\theta)}{tan(60+\theta).cot(60+\theta)}\\\\\\=\dfrac{1}{1}\\\\\\=1\end{gathered}

tan(60+θ).tan(30−θ)

cos

2

(45+θ)+cos

2

(45−θ)

=

tan(60+θ).cot(90−30+θ)

cos

(

45+θ)+sin

2

(90−45+θ)

=

tan(60+θ).cot(60+θ)

cos

2

(45+θ)+sin

2

(45+θ)

=

1

1

=1

Right Hand Side = Left Hand Side.

Hence Proved.