Question

plz answer this question​

Answer 1

Answer:

Given

In ∆ ABC

• right angled at C
• tan A= 1 /√3

To find

1. SinA.Cos B+ CosA.SinB
2. 1+ tan²A = Sec²A

Solution

We know that ,

$\sf{ \to \: tanA = \frac{1}{ \sqrt{3} } } \\ \\ \sf{ \to \: tanA = 30 \degree}$

Now,

⚡In an triangle the the sum of angles is 180°.

let us assume angle B as x°.

$\sf{ \to90 + 30 + x \degree = 180 \degree} \\ \sf{ \to120 \degree + x= 180 \degree} \\ \sf { \to \: x = 60 \degree}$

$\sf{\therefore \: \angle \: c \: is \: 60 \degree}$

Now subsitute the values

$\sf{Sin30 .Cos 60 + Cos30 .Sin60} \\ \\ \sf{ \to \: \frac{1}{2} \times \frac{1}{2} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2}} \\ \\ \sf{ \to \: \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1}$

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Q2

$\sf{tanA = \frac{1}{ \sqrt{3}} } \\ \\ \sf{ \to \: tanA = \frac{perpendicular}{base} } \\ \\ \sf{so\: perpendicular = 1units} \\ \sf{base = \sqrt{3} \: units}$

Apply phythagoras Theorem in ∆ ABC

$\sf{ \to \: AB²= AC²+ CB²} \\ \sf{ \to \: AB = 2units}$

Now,

L.H.S

$\sf{1 + {tan}^{2}A } \\ \sf{ \to \: 1 + ({ \frac{1}{ \sqrt{3} } })^{2} } \\ \sf { \to1 + \frac{1}{3} = \frac{4}{3} }$

R.H.S

$\sf{secA = \frac{h}{b} } \\ \sf{ \to \: secA = \frac{2}{ \sqrt{3} } } \\ \\ \sf{ \to {sec}^{2} A = { \frac{4}{3} }}$

therefore LHS= RHS

hence proved

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$