Question

Show that:
cos²(45° + ø) + cos²(45° - ø)/
tan(60° + ø) tan(30° - ø)
= 1​

Answer 1

SOLUTION:

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

Step-by-step explanation:

$\fbox { \underline {\underline{important \: points}}}$

:-- cosA = sin(90-A)

:-- tanA= cot(90-A)

:--cos²A + sin²A= 1

:-- tanA . cot A = 1

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L.H.S

$\to \frac{cos²(45+∅) + cos²(45-∅)}{tan(60- ∅) .tan(30-∅)} \\ \\ \to \frac{cos²(45+∅) + sin²(90 - 45 + ∅)}{tan(60+∅) .cot(90 - 30 + ∅)} \\ \\ \to \: \frac{cos²(45+∅) + sin²(45 + ∅)}{tan(60+∅) .cot(60+ ∅)} \\ \\ \to \: \frac{1}{1} = 1 = \bold{R.H.S}$

R.H.S

1

therefore LHS= RHS

hence proved