Question

PROVE THAT, cos² (45°+ø) + cos² (45°-ø)
tan (60° + ø) tan(30°- ø)=1
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Answer 1

Answer:

1

Step-by-step explanation:

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Answer 2

Given

[cos²(45°+ø) + cos²(45°-ø)]/[tan (60° + ø) tan(30°- ø)] = 1

To Prove

Prove that, [cos²(45°+ø) + cos²(45°-ø)]/[tan (60° + ø) tan(30°- ø)] = 1

L.H.S. = R.H.S.

$\rule{200}2$

Proof

Taking L.H.S.

⇒ [cos²(45°+ø) + cos²(45°-ø)]/[tan (60° + ø) tan(30°- ø)]

Used formulas: cosø = sin(90°-ø) and tanø = cot(90°-ø)

⇒ [sin²(90 -45°+ ø) + cos²(45°+ø)]/[tan(60°+ø). tan(30° + ø)]

⇒ [sin²(45°+ø) + cos²(45°+ø)]/[cot(90°-60°+ø). tan(30°+ø)]

We know that, sin²ø + cos²ø = 1

⇒ 1/[cot(30°+ø) tan(30°+ø)]

Also, 1/cotø = tanø

⇒ $\sf{\frac{1}{cot(30^{\circ} + \theta) \frac{1}{cot(30^{\circ}+ \theta)}}}$

⇒ 1/1

⇒ 1

L.H.S. = R.H.S

Hence, proved