Question

plzzzzzzz you answer me

Answer 1

Hey !!!

From given

1 - tan¢ /1 + tan¢ = tan45° - tan¢ /1 + 1×tan¢ {we can write }

= tan45° - tan¢ /1 - tan45°×tan¢

we know a formula tanA - tanB /1 + tanA×tanB = tan(A -B) like that

since , tan(45°-¢) = √3-1/√3+1

tan(45°-¢) = √3 -1 /√3+1×√3-1/√3-1 [multiplying by √3-1 on numerator and denominator]

we get .

tan(45°-¢) = (√3-1)²/(√3)²-(1)²

tan(45°-¢)= 3 +1 -2√3/2

4-2√3/2 = tan(45°-¢)

2(2-√3) /2 = tan(45°-¢)

2-√3 = tan(45°-¢)

tan15° = tan(45°-¢) •°• tan15° = 2-√3

15° = 45° -¢

15° - 45° = -¢

-30° = -¢

¢ = 30°

since , sin¢/sin2¢ = sin30°/cos60°

=> 1 /2/1/2 = 1 prooved

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Hope it helps you !!

@Rajukumar111