Question
If $A=30^{\circ}$; show that:$(\sin A-\cos A)^2=1-\sin 2 A$
✓
Answer
Given that $A =30^{\circ}$
$\text { LHS }=(\sin A -\cos A )^2$
$=\left(\sin 30^{\circ}-\cos 30^{\circ}\right)^2$
$=\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^2$
$=\frac{1}{4}+\frac{3}{4}-\frac{\sqrt{3}}{2}$
$=1-\frac{\sqrt{3}}{2}$
$=2-\frac{\sqrt{3}}{2}$
$\text{RHS} = 1 – \sin 2A$
$= 1 – \sin 2(30^\circ)$
$= 1 – \sin 60^\circ$
$=1-\frac{\sqrt{3}}{2}$
$=\frac{2-\sqrt{3}}{2}$
$\text { LHS }=\text { RHS }$
$\text { LHS }=(\sin A -\cos A )^2$
$=\left(\sin 30^{\circ}-\cos 30^{\circ}\right)^2$
$=\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)^2$
$=\frac{1}{4}+\frac{3}{4}-\frac{\sqrt{3}}{2}$
$=1-\frac{\sqrt{3}}{2}$
$=2-\frac{\sqrt{3}}{2}$
$\text{RHS} = 1 – \sin 2A$
$= 1 – \sin 2(30^\circ)$
$= 1 – \sin 60^\circ$
$=1-\frac{\sqrt{3}}{2}$
$=\frac{2-\sqrt{3}}{2}$
$\text { LHS }=\text { RHS }$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
More "[4 marks sum]" from Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]→Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] — all question types→MATHEMATICS STD 9 question papers→ICSE Board STD 9 question papers→ICSE Board question paper generator→
Similar questions
Evaluate the following :$(8.12)^3 - (3.12)^3$
→If $\log _2(x+y)=\log _3(x-y)=\frac{\log 25}{\log 0.2}$, find the values of $x$ and $y$.
→The cross-section of a 6 m long piece of metal is shown in the figure. Calculate:
(i) the area of the cross-section
(ii) the volume of the piece of metal in cubic centimetres.
(i) the area of the cross-section
(ii) the volume of the piece of metal in cubic centimetres.
Simplify the following$\frac{\sqrt{5}-2}{\sqrt{5}+2}-\frac{\sqrt{5}+2}{\sqrt{5}-2}$
→Calculate the amount and the compound interest on ₹ 25000 for 3 years, the rates of interest for the successive years being 8%, 9% and 10%, compounded annually.
Construct a$ \triangle RST$ with side $ST = 5.4 \ cm, \text{RST} = 60^\circ $ and the perpendicular from $R$ on $ST = 3.0 \ cm.$
→Solve the following:$\log _4 x+\log _4(x-6)=2$
→In the given figure, $AB > AC$. If BO and CO are the bisectors of $\angle B$ and $\angle C$ respectively, prove that $BO > CO$.
→Solve the following systems of equations by using the method of cross multiplication:
$\frac{5}{x+1}-\frac{2}{y-1}=\frac{1}{2}, \frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$, where $x \neq-1$ and $x \neq 1$
$\left[\right.$ Hint. Put $\frac{1}{x+1}=u$ and $\left.\frac{1}{y-1}=v\right]$
→$\frac{5}{x+1}-\frac{2}{y-1}=\frac{1}{2}, \frac{10}{x+1}+\frac{2}{y-1}=\frac{5}{2}$, where $x \neq-1$ and $x \neq 1$
$\left[\right.$ Hint. Put $\frac{1}{x+1}=u$ and $\left.\frac{1}{y-1}=v\right]$
Prove that:
$\left(\frac{x^a}{x^b}\right)^{(a+b-c)} \cdot\left(\frac{x^b}{x^c}\right)^{(b+c-a)} \cdot\left(\frac{x^c}{x^a}\right)^{c+a-b}=1$.
→$\left(\frac{x^a}{x^b}\right)^{(a+b-c)} \cdot\left(\frac{x^b}{x^c}\right)^{(b+c-a)} \cdot\left(\frac{x^c}{x^a}\right)^{c+a-b}=1$.