Question

In ΔABC if 3 Cot A = 4 find the value of sinA.cosC +cosA.sinC

Answer 1

Step-by-step explanation:

Given:-

In ΔABC , 3 Cot A = 4

To find :-

In ΔABC if 3 Cot A = 4 find the value of

sinA.cosC +cosA.sinC

Solution:-

Method : 1

See the attachment

Method:2-

Given that:

In ΔABC 3 Cot A = 4

=>3 Cot A = 4

=>Cot A = 4/3

We know that Cosec^2 A - Cot^2 A = 1

=>Cosec^2 A = 1 + Cot^2 A

=>Cosec^2 A = 1 + (4/3)^2

=>Cosec^2 A = 1+(16/9)

=>Cosec^2 A = (9+16)/9

=>Cosec^2 A = 25/9

=>Cosec A = √(25/9)

=>Cosec A = 5/3

We know that Sin A = 1 / Cosec A

=> Sin A = 1/(5/3)

Sin A = 3/5

We know that

Sin^2 A + Cos^2 A = 1

=>(3/5)^2 + Cos^2 A = 1

=>(9/25) + Cos^2 A = 1

=>Cos^2 A = 1-(9/25)

=>Cos^2 A = (25-9)/25

=> Cos^2 A = 16/25

=> Cos A = √(16/25)

=> Cos A = 4/5

Now

Sin C = Opposite side to angle C/ Hypotenuse

=>Sin C = 4/5

Cos C = Adjacent side to angle C/Hypotenuse

=> Cos C = 3/5

Now,

sinA.cosC +cosA.sinC

=>(3/5)×(3/5) + (4/5)×(4/5)

=>(9/25) + (16/25)

=>(9+16)/25

=>25/25

=>1

Answer:-

the value of sinA.cosC +cosA.sinC for the given problem is 1