Question

If cos A= 12/13 and cot B =24/7 ,A lies on the second Quartdant and B in third quadrant. Find sum (a+b) ,Tan(a+b),cos(a+b)​ I will do as Brain least

Answer 1

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.

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 .