Question

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Answer 1

Question :

Prove that $\sf{2sin^{2}\dfrac{3\pi}{4} + 2cos^{2}\dfrac{\pi}{4} + 2sec^{2}\dfrac{\pi}{3} = 10}$

To find :

• To Prove that LHS is equal to the RHS of the equation.

$\sf{2sin^{2}\dfrac{3\pi}{4} + 2cos^{2}\dfrac{\pi}{4} + 2sec^{2}\dfrac{\pi}{3} = 10}$

Knowledge required :

• The value of $\sf{\pi = 180^{\circ}}$

• $\sf{sin(\alpha + \beta) = sin\alpha cos\beta + cos\alpha sin\beta}$

• sin90° = 1
• cos90° = 0
• sin45° = cos45° = 1/√2
• sec60° = 2

Solution :

From the given equation, we get that :

• LHS = $\sf{2sin^{2}\dfrac{3\pi}{4} + 2cos^{2}\dfrac{\pi}{4} + 2sec^{2}\dfrac{\pi}{3}}$

• RHS = $\sf{10}$

Let's solve the LHS of the equation,

$:\implies \sf{2sin^{2}\dfrac{3\pi}{4} + 2cos^{2}\dfrac{\pi}{4} + 2sec^{2}\dfrac{\pi}{3}}$

Putting $\sf{\pi = 180^{\circ}}$, we get :

$:\implies \sf{2sin^{2}\bigg[\dfrac{3(180^{\circ})}{4}\bigg] + 2cos^{2}\bigg[\dfrac{(180^{\circ})}{4}\bigg] + 2sec^{2}\bigg[\dfrac{(180^{\circ})}{3}\bigg]} \\ \\ :\implies \sf{2sin^{2}\bigg[\dfrac{(540^{\circ})}{4}\bigg] + 2cos^{2}\bigg[\dfrac{(180^{\circ})}{4}\bigg] + 2sec^{2}\bigg[\dfrac{(180^{\circ})}{3}\bigg]} \\ \\ :\implies \sf{2sin^{2}(135^{\circ}) + 2cos^{2}(45^{\circ}) + 2sec^{2}(60^{\circ})}$

By taking 2 common in the equation, we get :

$:\implies \sf{2[sin^{2}(135^{\circ}) + cos^{2}(45^{\circ}) + sec^{2}(60^{\circ})]}$

We know that $\sf{sin^{2}(135^{\circ})}$ can be written as $\sf{sin(135^{\circ})sin(135^{\circ})}$, so by substituting it in the equation, we get :

$:\implies \sf{2[sin(135^{\circ})sin(135^{\circ}) + cos^{2}(45^{\circ}) + sec^{2}(60^{\circ})]} \\ \\ :\implies \sf{2[sin(90^{\circ} + 45^{\circ})sin(90^{\circ} + 45^{\circ}) + cos^{2}(45^{\circ}) + sec^{2}(60^{\circ})]}\:\:\:\:[\because \sf{sin135^{\circ} = sin(90^{\circ} + 45^{\circ})}$

To find the value sin(90° + 45°) :

$:\implies \sf{sin(90^{\circ} + 45^{\circ}) = sin(90^{\circ})cos(45^{\circ}) + cos(90^{\circ})sin(45^{\circ})} \\ \\ :\implies \sf{sin(90^{\circ} + 45^{\circ}) = (1)\bigg(\dfrac{1}{\sqrt{2}}\bigg) + (0\bigg(\dfrac{1}{\sqrt{2}}\bigg)} \\ \\ :\implies \sf{sin(90^{\circ} + 45^{\circ}) = \dfrac{1}{\sqrt{2}}} \\ \\ \boxed{\therefore \sf{sin(90^{\circ} + 45^{\circ}) = \dfrac{1}{\sqrt{2}}}}$

Hence the value of $\sf{sin(90^{\circ} + 45^{\circ})}$ is $\sf{\dfrac{1}{\sqrt{2}}}$

Now substituting the values of $\sf{sin(90^{\circ} + 45^{\circ})}$, we get :

$:\implies \sf{2\bigg[\dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{2}} + \dfrac{1}{(\sqrt{2})^{2}} + (2)^{2}\bigg]} \\ \\ :\implies \sf{2\bigg[\dfrac{1}{(\sqrt{2})^{2}} + \dfrac{1}{(\sqrt{2})^{2}} + (2)^{2}\bigg]} \\ \\ :\implies \sf{2\bigg(\dfrac{1}{2} + \dfrac{1}{2} + 4\bigg)} \\ \\ :\implies \sf{2\bigg(\dfrac{1 + 1 + 8}{2}\bigg)} \\ \\ :\implies \sf{2\bigg(\dfrac{10}{2}\bigg)} \\ \\ :\implies \sf{\not{2}\bigg(\dfrac{10}{\not{2}}\bigg)} \\ \\ :\implies \sf{10} \\ \\ \boxed{\therefore \sf{2sin^{2}\dfrac{3\pi}{4} + 2cos^{2}\dfrac{\pi}{4} + 2sec^{2}\dfrac{\pi}{3} = 10}}$

Now putting the LHS and RHS together, we get :

$:\implies \sf{LHS = RHS} \\ \\ :\implies \sf{10 = 10}$

Thus,

$\sf{2sin^{2}\dfrac{3\pi}{4} + 2cos^{2}\dfrac{\pi}{4} + 2sec^{2}\dfrac{\pi}{3} = 10}$

Proved !!

Answer 2

To prove:

$\sf{2\sin² \dfrac{3\pi}{4}+ 2\cos² \dfrac{\pi}{4}+2\sec² \dfrac{\pi}{3} = 10}$

Proof:

$\sf{➪\:LHS = 2\sin² \dfrac{3\pi}{4}+ 2\cos² \dfrac{\pi}{4}+2\sec² \dfrac{\pi}{3}}$

$\sf{➪\:LHS = 2\sin² \dfrac{3(180°)}{4}+ 2\cos² \dfrac{(180°)}{4}+2\sec² \dfrac{(180°)}{3}}$

$\sf{➪\:LHS = 2\sin² 135° + 2\cos²45°+2\sec² 60°}$

$\sf{➪\:LHS = 2\sin²(180° -45°)+ 2\cos²45°+2\sec² 60°}$

$\sf{➪\:LHS = 2\sin² 45° + 2\cos²45°+2\sec²60°}$

$\sf{➪\:LHS = 2\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2} + 2\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2}+2(2)²}$

$\sf{➪\:LHS = 1 + 1 +8}$

$\sf{➪\:LHS = 10\:➪}$ $\bold{\red{LHS = RHS}}$

$\large{\bold{★\:Hence\:proved}}$

Table to remember:

$\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered}$