Question

Prove that $\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\\!\\!\pi\\!\\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{3}}\text{=}\dfrac{\text{3}}{\text{2}}$?

Answer 1

For the given question we have to prove LHS=RHS. Let us take LHS and start solving. First of all we have to reduce the terms which are >180 to <=180. Then we have to put values in the equation and therefore by solving the remaining part we can prove LHS=RHS.

Complete step-by-step solution:
For the given question we are given to prove the given equation. For solving this let us assume the whole equation as ‘S’ and then consider it as equation (1).
Let us assume the given equation as ‘S’.
$\text{S=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\\!\\!\pi\\!\\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{3}}\text{=}\dfrac{\text{3}}{\text{2}}$
Let us consider the above equation as equation (1).
$\text{S=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\\!\\!\pi\\!\\!\text{ }}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{ }\\!\\!\pi\\!\\!\text{ }}{\text{3}}\text{=}\dfrac{\text{3}}{\text{2}}........\left( \text{1} \right)$
As we know $\pi ={{180}^{\circ }}$, let us consider it as equation (2).
$\pi ={{180}^{\circ }}..........\left( 2 \right)$
Taking LHS( left hand side ) to solve the problem.
By substituting equation (2) in equation (1), we get
$\Rightarrow \text{S=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{18}{{\text{0}}^{\text{o}}}}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\\!\\!\times\\!\\!\text{ 18}{{\text{0}}^{\text{o}}}}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{18}{{\text{0}}^{\text{o}}}}{\text{3}}$
Let us consider the above equation as equation (3).
$\Rightarrow \text{S=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{18}{{\text{0}}^{\text{o}}}}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\dfrac{\text{7 }\\!\\!\times\\!\\!\text{ 18}{{\text{0}}^{\text{o}}}}{\text{6}}\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{18}{{\text{0}}^{\text{o}}}}{\text{3}}..........\left( \text{3} \right)$
We can write $\dfrac{7\times 180}{6}$as$180+\dfrac{180}{6}$, so rewrite the above equation by replacing the term$\dfrac{7\times 180}{6}$.
$\Rightarrow \text{S=2si}{{\text{n}}^{\text{2}}}\dfrac{\text{18}{{\text{0}}^{\text{o}}}}{\text{6}}\text{+cose}{{\text{c}}^{\text{2}}}\left( \text{180+}\dfrac{\text{180}}{\text{6}} \right)\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{18}{{\text{0}}^{\text{o}}}}{\text{3}}$
As $\theta$lies in the third quadrant, $\theta$ will be negative.
$\Rightarrow S=2{{\sin }^{2}}\dfrac{{{180}^{\circ }}}{6}+\text{cose}{{\text{c}}^{\text{2}}}\left( -\dfrac{180}{6} \right){{\cos }^{2}}\dfrac{{{180}^{\circ }}}{3}$
By simplifying the equation (3) a bit, we get
$\Rightarrow S=2{{\sin }^{2}}\dfrac{{{180}^{\circ }}}{6}+\text{cose}{{\text{c}}^{2}}-\dfrac{180}{6}{{\cos }^{2}}\dfrac{{{180}^{\circ }}}{3}$
Let us consider the above equation as equation (4).
$\Rightarrow S=2{{\sin }^{2}}\dfrac{{{180}^{\circ }}}{6}+\text{cose}{{\text{c}}^{2}}-\dfrac{180}{6}{{\cos }^{2}}\dfrac{{{180}^{\circ }}}{3}.......\left( 4 \right)$
Substituting the values in the equation (4), we get
$\Rightarrow S=2{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( -2 \right)}^{2}}\times {{\left( \dfrac{1}{2} \right)}^{2}}$
By simplifying a bit,
$\Rightarrow S=2\times \dfrac{1}{4}+1$
$\Rightarrow \dfrac{3}{2}$
Let us consider the above value as equation (5).
$\Rightarrow S=\dfrac{3}{2}................\left( 5 \right)$
$\Rightarrow \text{RHS}$
Hence proved.
Therefore from the equation (5) we can say that $\text{LHS=RHS}$.

Note: We can see the term $\dfrac{7\pi }{6}$ which can be written as $\left( \pi +\dfrac{\pi }{6} \right)$ we know that this will lie in quadrant 3 and where cosec is negative here but in the question we can see a square for cosec so this will be positive, so here negative sign is not effective. So, be careful while reducing angles.