Question

In ΔABC, right angled at B. If tan A = `1/sqrt3` , find the value of

1.  sin A cos C + cos A sin C
2. cos A cos C − sin A sin C

If ΔABC, ∠B = 90° and Tan A = `1/sqrt(3)`. Prove that

1. Sin A. cos C + cos A. Sin c = 1
2. cos A. cos C - sin A. sin C = 0

Answer 1

tan A = `1/sqrt3`

`"BC"/"AB"=1/sqrt3`

If BC is k, then AB will be `sqrt3k`, where k is a positive integer.

In ΔABC,

AC² = AB² + BC²

= `(sqrt3k)^2 + (k)^2`

= 3k² + k² 

= 4k²

∴ AC = 2k

sin A = `("Side adjacent to ∠A")/"Hypotenuse" = ("BC")/("AC") = k/(2k) = 1/2`

cos A = `("Side adjacent to ∠A")/"Hypotenuse" = ("AB")/("AC") = (sqrt3k)/(2k) = sqrt3/2`

sin C = `("Side adjacent to ∠C")/"Hypotenuse" = ("AB")/("AC") = (sqrt3k)/(2k) = sqrt3/2`

cos C = `("Side adjacent to ∠C")/"Hypotenuse" = ("BC")/("AC") = (k)/(2k) = 1/2`

(i) sin A cos C + cos A sin C

= `(1/2)(1/2)+(sqrt3/2)(sqrt3/2) `

= `1/4 + 3/4`

= `4/4`

= 1

(ii) cos A cos C − sin A sin C

= `(sqrt3/2)(1/2)-(1/2)(sqrt3/2)`

= `sqrt3/4 - sqrt3/4`

= 0

Answer 2

In ΔABC, ∠B = 90°,

As, tan A = `1/sqrt(3)`

⇒ `("BC")/("AB") = 1/sqrt(3)`

Let BC = x and AB = x = `sqrt(3)`

Using Pythagoras the get

AC = `sqrt("AB"^2 + "BC"^2)`

= `sqrt((xsqrt(3))^2 + x^2)`

= `sqrt(3x^2 + x^2)`

= `sqrt(4x^2)`

= 2x

Now,

(i) LHS = sin A. cos C + cos A . sin C

= `("BC")/("AC") . ("BC")/("AC") + ("AB")/("AC") .("AB")/("AC")`

= `(("BC")/("AC"))^2 + (("AB")/("AC"))^2`

= `(x/(2x))^2 + ((xsqrt(3))/(2x))^2`

= `1/4 +3/4`

= 1

= RHS

(ii) LHS = cos A . cos C - sinA . sinC

= `("AB")/("AC") .("BC")/("AC") -("BC")/("AC") .("AB")/("AC")`

= `(xsqrt(3))/(2x) .x/2x - x/2x.(xsqrt(3))/(2x)`

= `sqrt(3)/4 - sqrt(3)/4`

= 0

= RHS