Question

kindly solve que 24
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Answer 1

Analysis→

Here we're given an equation tan²30°sin30°+cos60°sin²90°tan²60°-2tan45°cos²0°sin90°. And we've to evaluate the equation and find its value. And we know that :-

$\dashrightarrow\rm{Sin30°=\dfrac{1}{2}}$

$\dashrightarrow\rm{Sin90°=1}$

$\dashrightarrow\rm{Cos0°=1}$

$\dashrightarrow\rm{Cos60°=\dfrac{1}{2}}$

$\dashrightarrow\rm{Tan30°=\dfrac{1}{\sqrt{3}}}$

$\dashrightarrow\rm{Tan45°=1}$

$\dashrightarrow\rm{Tan60°={\sqrt{3}}}$

Given→

• Equation= tan²30°sin30°+cos60°sin²90°tan²60°-2tan45°cos²0°sin90°.

To Find→

The value of the equation.

Answer→

$\rm{tan^230°sin30°+cos60°sin^290°tan^260°-2tan45°cos^20°sin90°}$

$\displaystyle{\rm{ \Big( \dfrac{1}{\sqrt{3}} \Big)^2. \Big( \dfrac{1}{2} \Big)+ \Big( \dfrac{1}{2} \Big).(1)^2.\big( {\sqrt{3}} \big)^2-2 \big[(1)^2.(1)^2.1 \big]}}$

$\displaystyle{\rm{ \Big(\dfrac{1}{3}\times\dfrac{1}{2}\Big)+\Big(\dfrac{1}{2}\times1\times3\Big)-2(1\times1\times1)}}$

$\displaystyle{\rm{ \Big(\dfrac{1}{3}\times\dfrac{1}{2}\Big)+\Big(\dfrac{1}{2}\times1\times3\Big)-(2)}}$

$\implies\rm{ \Big(\dfrac{1}{6}\Big) +\Big(\dfrac{1}{2}\times3\Big)-(2)}$

$\implies\rm{ \Big(\dfrac{1}{6}\Big)+\Big(\dfrac{3}{2}\Big) -(2)}$

Taking the LCM[LCM(1,2,6)=6]:-

$\implies\rm{ \dfrac{1}{6}+\dfrac{9}{6}-\dfrac{12}{6}}$

$\implies\rm{ \dfrac{1+9-12}{6}}$

$\implies\rm{ \dfrac{10-12}{6}}$

$\implies\rm{ \dfrac{-2}{6}}$

$\implies\rm{-\dfrac{\cancel{2}}{\cancel{6}}}$

$\implies\rm{-\dfrac{1}{3}}$

${\boxed{\boxed{\implies{\bf{-\dfrac{1}{3}\checkmark}}}}}$

☞ Hence the value of the given equation is $\rm{-\dfrac{1}{3}}$ which is the required answer.

HOPE IT HELPS.

Answer 2

Answer:

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