Question

Mark the correct option choosing the correct coordinates.
A. X-coordiate In a right angled triangle ABC, right angled at B, the ratio of AB to AC is
$1:\sqrt{2}$, then $\frac{4tanA}{1+ta{n}^{2}A}$
y-coordinate In a right angled triangle ABC, right angled at B, $\angle ACB=\theta$, AB=2 cm and BC=1 cm, then $2(si{n}^2\theta +co{s}^2\theta )$.

B. x-coordinate If$sinB=\frac{1}{2}$, then value of $3cosB-4co{s}^{3}B$.
y-coordinate the value of $ta{n}^2\theta +\frac{1}{ta{n}^2{\theta }^{\prime }}$
if $tan\theta +\frac{1}{tan\theta }=\sqrt{2}$

C. x-coordinate The value of
$(\frac{-1}{tanA}-\frac{sinA}{1+cosA})$, if $cosecA=1$,
y-coordinate If $cos\theta =\frac{3}{5}$, then -$(cot\theta +cosec\theta )$

• A. C is the reflection of A in B.
  
• B. C is the reflection of A on x- axis
  
• C. A is the reflection of C on y-axis
  
• D. None of the above

Answer 1

The correct option is A

C is the reflection of A in B.

(A) x- coordinate


Give $B=1,H=\sqrt{2}$
$\therefore p=\sqrt{H^2-B^2}$
(by pythagoras theorem)
$=\sqrt{2-1}=1\Rightarrow tanA=\frac{P}{B}=\frac{1}{1}\therefore \frac{4tanA}{1+ta{n}^{2}A}=\frac{1\times 1}{1+1}=2$
y-coordinate

Now $Ac=\sqrt{A{B}^2+B{C}^2}=\sqrt{4+1}=\sqrt{5}\therefore sin\theta =\frac{2}{\sqrt{5}}andcos\theta =\frac{1}{\sqrt{5}}$
Now,
$si{n}^2\theta +co{s}^2\theta =\frac{2^2}{5}+\frac{1^2}{(\sqrt{5}{)}^2}=\frac{4}{5}+\frac{1}{5}=\frac{5}{5}=1\therefore 2(si{n}^2\theta +co{s}^2\theta )=2$
$\Rightarrow$ $A(2,2)$

(B) x-coordinate
Given,
$sinB=\frac{1}{2}\Rightarrow cosB=\frac{\sqrt{3}}{2}\therefore 3cosB=4co{s}^{2}B=3\times \frac{\sqrt{3}}{2}-4{\left(\frac{\sqrt{3}}{2}\right)}^3=\frac{3\sqrt{3}}{2}-4\frac{3\sqrt{3}}{8}=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}=0$
y-coordinate
Given $ta{n}^2\theta +\frac{1}{ta{n}^2\theta }$
Now, $(tan\theta +\frac{1}{tan\theta })=ta{n}^2\theta +\frac{1}{ta{n}^2\theta }+2tan\theta \times \frac{1}{tan\theta }\Rightarrow (\sqrt{2}{)}^2=ta{n}^2\theta +\frac{1}{ta{n}^2\theta }+2\Rightarrow ta{n}^2\theta +\frac{1}{ta{n}^2\theta }=0\therefore B(0,0)$

(C) x-coordinate
Given $cosecA=2$
$\Rightarrow sinA=\frac{1}{2}\Rightarrow cosA=\frac{\sqrt{3}}{2}andtanA=\frac{1}{\sqrt{3}}\therefore -(\frac{1}{tanA}+\frac{sinA}{1+cosA})=-(\sqrt{3}+\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}})$
$=-(\sqrt{3}+\frac{1}{2+\sqrt{3}})=-\frac{(2\sqrt{3}+3+1)}{2+\sqrt{3}}=-\frac{(2\sqrt{3}+4)}{2+\sqrt{3}}=-\frac{2(2+\sqrt{3})}{2+\sqrt{3}}=-2$

y-coordinate
Given $cos\theta =\frac{3}{5}\therefore sin\theta =\frac{4}{5}\Rightarrow cot\theta =\frac{3}{4}andcosec\theta =\frac{5}{4}$
$\therefore -(cot\theta +cosec\theta )=-(\frac{3}{4}+\frac{5}{4})$
$=-\left(\frac{8}{4}\right)=-2\therefore C(-2,-2)$
Hence, C is the reflection of A in B (origin)