Question

If cos A = $\frac{7}{25}$ and $\frac{3\pi}{2}$ < A < 2π, then find the value of cot $(\frac{A}{2})$.

Answer 1

cosA = 7/25 where 3π/2 < A < 2π

we know , cos2x = (1 - tan²x)/(1 + tan²x)

so, cosA = {1 - tan²(A/2)}/{1 + tan²(A/2)}

7/25 = {1 - 1/cot²(A/2)}/{1 +1/cot²(A/2)}

7/25 = {cot²(A/2) - 1}/{cot²(A/2) + 1}

7{cot²(A/2) + 1} = 25{cot²(A/2) - 1}

7cot²(A/2) + 7 = 25cot²(A/2) - 25

7 + 25 = 25cot²(A/2) - 7cot²(A/2)

32 = 18cot²(A/2)

cot²(A/2) = 16/9

cot(A/2) = ±4/3

but 3π/2 < A < 2π or, 3π/4 < (A/2) < π ,
e.g., (A/2) lies in 2nd quadrant .

so, cot(A/2) = -4/3

Answer 2

HELLO DEAR,



GIVEN:-
cosA = 7/25

we know:- cos2x = (1 - tan²x)/(1 + tan²x)

so, cosA = {1 - tan²(A/2)}/{1 + tan²(A/2)}

7/25 = {1 - 1/cot²(A/2)}/{1 +1/cot²(A/2)}

7/25 = {cot²(A/2) - 1}/{cot²(A/2) + 1}

7{cot²(A/2) + 1} = 25{cot²(A/2) - 1}

7cot²(A/2) + 7 = 25cot²(A/2) - 25

7 + 25 = 25cot²(A/2) - 7cot²(A/2)

32 = 18cot²(A/2)

cot²(A/2) = 16/9

cot(A/2) = ±4/3

[but 3π/2 < A < 2π or, 3π/4 < (A/2) < π ,]

so, cot(A/2) = -4/3



I HOPE IT'S HELP YOU DEAR,
THANKS