Question

In $\Delta ACB$ right-angled at B, if AB =3 units, BC = 4units and $\angle ACB=\theta ,$ then the value of $[cose{c}^2\theta -ta{n}^2\theta +co{s}^2\theta +se{c}^2\theta -co{t}^2\theta +2-(-si{n}^2\theta \left)\right]isequalto$

• A.
  $\frac{5}{2}$
• B.
  $\frac{3}{4}$
• C.
  5
• D.
  1

Answer 1

The correct option is C

5
Given: AB = 3 units, BC = 4 units and $\angle ACB=\theta$

In $\Delta ABC,$ By pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2\Rightarrow A{C}^2=3^2+4^2\Rightarrow AC=\sqrt{25}=5unitsNow,Sin\theta =\frac{AB}{AC}=\frac{3}{5},cos\theta =\frac{BC}{AC}=\frac{4}{5},tan\theta =\frac{AB}{BC}=\frac{3}{4}.cosec\theta =\frac{1}{sin\theta }=\frac{5}{3},sec\theta =\frac{1}{cos\theta }=\frac{5}{4},cot\theta =\frac{1}{tan\theta }=\frac{4}{3}$
Substituting these values in $[cose{c}^2\theta -ta{n}^2\theta +co{s}^2\theta +se{c}^2\theta -co{t}^2\theta +2-(-si{n}^2\theta \left)\right]$

we get,

$[{\left(\frac{5}{3}\right)}^2-{\left(\frac{3}{4}\right)}^2+{\left(\frac{4}{5}\right)}^2+{\left(\frac{5}{4}\right)}^2-{\left(\frac{4}{3}\right)}^2+2+{\left(\frac{3}{5}\right)}^2]=\frac{25}{9}-\frac{9}{16}+\frac{16}{25}+\frac{25}{16}-\frac{16}{9}+2+\frac{9}{25}=\frac{25-16}{9}+\frac{16+9}{25}+\frac{25-9}{16}+2=\frac{9}{9}+\frac{25}{25}+\frac{16}{16}+2=1+1+1+2=5$

Hence, the correct answer is option c.