Question

Q). In Triangle ABC , right- angled at B , if tan A =
$\frac{1}{ \sqrt{3} }$
Find the value of :
1). sin A cos C + cos A sin C
2). cos A cos C - sin A sin C ​

Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

In ∆ ABC right angled at B

${\tt{tanA = \frac{1}{ \sqrt{3} } }}$

${\bf{\blue{\underline{To:Find:}}}}$

$\star{\tt{ \: SinA \: CosC + CosA \: SinC}}$

$\star{\tt{ \: CosA\:CosC -SinA \: SinC}}$

${\bf{\blue{\underline{Now:}}}}$

In ∆ABC,

$\implies{\tt{ {h}^{2} = {b}^{2} + {p}^{2} }}$

$\implies{\tt{ {h}^{2} = {( \sqrt{3} )}^{2} + {(1)}^{2} }}$

$\implies{\tt{ {h}^{2} = 3 + 1 }}$

$\implies{\tt{ {h}^{2} = 4 }}$

$\implies{\boxed{ \purple{\tt{ h = \sqrt{4} }}}}$

$\star{ \boxed{\tt{ tanA = \frac{1}{ \sqrt{3} } = \frac{p}{b} }}}$

$\star{ \boxed{\tt{ cosA = \frac{ \sqrt{3} }{ 2 } = \frac{b}{h} }}}$

$\star{ \boxed{\tt{ sinA= \frac{ 1 }{ 2 } = \frac{p}{b} }}}$

$\star{ \boxed{\tt{ cosC = \frac{ 1 }{ 2 } = \frac{b}{h} }}}$

$\star{ \boxed{\tt{ sinC = \frac{ \sqrt{3} }{ 2 } = \frac{p}{b} }}}$

Now put the value of these in given equations,

$(2).\implies{\tt{ \frac{1}{2} \times \frac{1}{2} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} }}$

$\implies{\tt{ \frac{1}{4} + \frac{3}{4} }}$

$\implies{\tt{ \frac{1 + 3}{4} }}$

$\implies{\tt{ \frac{4}{4} }}$

$\implies\boxed{\purple{\tt{ 1 \: Ans }}}$

$(2).\implies{\tt{ \frac{ \sqrt{3} }{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{ \sqrt{3} }{2} }}$

$\implies{\tt{ \frac{ \sqrt{3} }{4} - \frac{ \sqrt{3} }{4} }}$

$\implies{\tt{ \frac{ \sqrt{3} - \sqrt{3} }{4} }}$

$\implies{\tt{ \frac{ 0}{4} }}$

$\implies \boxed{\purple {\tt{ 0 \: Ans }}}$

Answer 2

GIVEN:

• In ∆ABC right angle at B
• tanA = 1/√3

TO FIND:

1. sinA.cosC + cosAsinC
2. cosAcosC - sinAsinC

If tanA = 1/√3

A = 30°

Because , tan30° = 1/√3 & by comparing A = 30°

Finding Angle C using angle sum property

Angle A + angle B + angle C = 180°

30° + 90° + angle C = 180°

angle = 180° - 120°

Angle C = 60°

SOLUTION 1:

sinAcosC + cosAsinC

It is in the form of sinAsinC + cosAsinC = sin(A +C)

Substituting the values of Angle A & C

→ sin( 30° + 60° )

→ sin90° = 1

SOLUTION 2:

sinAcosC + cosAsinC

It is in the form of cosAcosC - sinAsinC = cos(A+C)

Substituting the values of Angle A & C

→ cos(30° + 60°)

→ cos90° = 0

Answers

★ sinA.cosC + cosAsinC = 1

★ cosAcosC - sinAsinC = 0