In a ABC - - B -90- and tan A -13- Prove thati sin A -cos C-cos A -sin C-1ii cos A -cos C -sin A -sin C -0

In nbsp;∆ABC, nbsp;∠B=90°,As, nbsp;tanA=13⇒BCAB=13Let nbsp;BC=x nbsp;and nbsp;AB=x3Using nbsp;Pythagoras nbsp;theorem, nbsp;we nbsp;getAC= ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

In a ABC , ∠ ...

Question

In a $∆$ ABC, $∠$ B = 90 $°$ and tanA = $\frac{1}{\sqrt{3}}$ . Prove that
(i) sinA $\cdot$ cosC + cosA $\cdot$ sinC = 1 (ii) cosA $\cdot$ cosC $-$ sinA $\cdot$ sinC = 0

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Solution

$In ∆ ABC, ∠ B = 90 °, As, tanA = \frac{1}{\sqrt{3}} \Rightarrow \frac{BC}{AB} = \frac{1}{\sqrt{3}} Let BC = x and AB = x \sqrt{3} Using Pythagoras theorem, we get AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{{(x \sqrt{3})}^{2} + x^{2}} = \sqrt{3 x^{2} + x^{2}} = \sqrt{4 x^{2}} = 2 x$

Now,

$(i) LHS = sinA \cdot cosC + cosA \cdot sinC = \frac{BC}{AC} \cdot \frac{BC}{AC} + \frac{AB}{AC} \cdot \frac{AB}{AC} = {(\frac{BC}{AC})}^{2} + {(\frac{AB}{AC})}^{2} = {(\frac{x}{2 x})}^{2} + {(\frac{x \sqrt{3}}{2 x})}^{2} = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1 = RHS$

$(ii) LHS = cosA \cdot cosC - sinA \cdot sinC = \frac{AB}{AC} \cdot \frac{BC}{AC} - \frac{BC}{AC} \cdot \frac{AB}{AC} = \frac{x \sqrt{3}}{2 x} . \frac{x}{2 x} - \frac{x}{2 x} . \frac{x \sqrt{3}}{2 x} = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 = RHS$

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In a ∆ABC, ∠B = 90° and tanA = 13. Prove that (i) sinA·cosC + cosA·sinC = 1 (ii) cosA·cosC - sinA·sinC = 0

In ∆ABC, ∠B=90°,As, tanA=13⇒BCAB=13Let BC=x and AB=x3Using Pythagoras theorem, we getAC=AB2+BC2=x32+x2=3x2+x2=4x2=2x Now, i LHS=sinA·cosC+cosA·sinC=BCAC·BCAC+ABAC·ABAC=BCAC2+ABAC2=x2x2+x32x2=14+34=44=1=RHS ii LHS=cosA·cosC-sinA·sinC=ABAC·BCAC-BCAC·ABAC=x32x.x2x-x2x.x32x=34-34=0=RHS

In a $∆$ ABC, $∠$ B = 90 $°$ and tanA = $\frac{1}{\sqrt{3}}$ . Prove that
(i) sinA $\cdot$ cosC + cosA $\cdot$ sinC = 1 (ii) cosA $\cdot$ cosC $-$ sinA $\cdot$ sinC = 0