Question

Easy question, but still m bad at calculations.

If cos A = -12/13 and cot B = 24/7 , where A lies in second quadrant and B in third quadrant , find the value of sin ( A + B )?

I need steps

Answer 1

Answer:

-36/325

Step-by-step explanation:

sin(A+B) = sinA.cosB + cosA.sinB

by using triangle, we can calculate the values that v need using the given data

sin^2A + cos^2A = 1

=> sinA = 5/13

and sinB = -7/25 , cosB = -24/25

now, Sin(A+B) = (5/13)*(-24/25) + (-12/13)*(-7/25)

= (-120/325) + (84/325)

= -36/325

signs are based on quadrants

hope it helps uh.....

inbox me...if any doubt....or it is wrong

Answer 2

Answer:

A+B = cos^-1(253/325)

tan(A+B) = 204/253

cos(A+B) = 253/325

Step-by-step explanation:

we have, cosA = 12/13

and cotB = 24/7

as, it is given that,A lies on 2nd quadrant and B lies on 3rd quadrant

so, cosA = -12/13

and cotB = 24/7

then, A = cos^-1(-12/13) = cos^-1(12/13)

B = cot^-1(24/7)

so, we have to calculate,,

A+B = cos^-1(12/13)+cot^-1(24/7) .... (1)

now, let cot^-1(24/7) = x

cotx = 24/7

cosecx = √{1+24²/7²}

= √{625/49}

= 25/7

so, sinx = 7/25

so, cosx = √{1-49/625}

cosx = √{576/625}

x = cos^-1(24/25)

by putting it in eq. 1

A+B = cos^-1(12/13)+cos^-1(24/25)

as, cos^-1(x)+cos^-1(y) = cos^-1{xy-√(1-x²)(1-y²)}

so, A+B,

cos^-1{12/13×24/25-√(25/169)(49/625)}

cos^-1{288/325-35/325} => A+B = cos^-1(253/325) ....ans.

and tan(A+B) =

let A+B= cos^-1(253/325) = y

cosy = 253/325

siny = √{1-64009/105625}

= √{41616/105625}

= 204/325

so, tany = (204/325)/(253/325)

= 204/253

y = tan^-1(204/253)

so, now..

tan{tan^-1(204/253)}

=> tan(A+B) => 204/253 .... ans.

and cos(A+B)

so, cos{cos^-1(253/325)} => cos(A+B) =>253/325 .... ans.