Question

Prove that :-

$\sf {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}\ =\ - \dfrac{1}{2}}$

Answer 1

Step-by-step explanation:

$\underline{\underline{\maltese\: \: \textbf{\textsf{To\ Prove}}}}$

$\sf \circ\ {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}\ =\ - \dfrac{1}{2}}$

$\underline{\underline{\maltese\: \: \textbf{\textsf{Solution}}}}$

➤ By taking the L.H.S

➠ That is,

$\sf \dashrightarrow {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}}$

• $\sf {Putting\ {\pi}\ =\ 180^{\circ}}$

$\sf \dashrightarrow {sin^2\ \dfrac{180^{\circ}}{6}\ +\ cos^2\ \dfrac{180^{\circ}}{3}\ -\ tan^2\ \dfrac{180^{\circ}}{6}}$

$\sf \dashrightarrow {sin^2\ 30^{\circ}\ +\ cos^2\ 60^{\circ}\ -\ tan^2\ 45^{\circ}}$

$\sf \dashrightarrow {(sin\ 30^{\circ})^2\ +\ (cos\ 60^{\circ})^2\ -\ (tan\ 45^{\circ})^2}$

➤ By putting,

• $\sf {sin\ 30^{\circ}\ =\ 1/2}$
• $\sf {cos\ 60^{\circ}\ =\ 1/2}$
• $\sf {tan\ 45^{\circ}\ =\ 1}$

$\sf \dashrightarrow {\bigg(\dfrac{1}{2} \bigg)^2\ +\ \bigg(\dfrac{1}{2} \bigg)^2\ -\ 1^2}$

$\sf \dashrightarrow {\dfrac{1}{4}\ +\ \dfrac{1}{4}\ -\ 1}$

$\sf \dashrightarrow {\dfrac{1\ +\ 1\ -\ 4}{4}}$

$\sf \dashrightarrow {\dfrac{2\ -\ 4}{4}}$

$\sf \dashrightarrow {\dfrac{-2}{4}}$

$\sf \dashrightarrow {\dfrac{-1}{2}}$

$\sf \dashrightarrow {R.H.S}$

$\large {\therefore{\boxed{\boxed{\sf {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}\ =\ - \dfrac{1}{2}}}}}}$

$\underline{\textbf{\textsf{Hence\ Proved...!!!}}}$

Answer 2

$\purple\bigstar$Given that :-

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• $\sf \circ\ {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}\ =\ - \dfrac{1}{2}}$

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$\orange\bigstar$Solution :-

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$\red\bigstar$ Taking L . H . S

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➽ we have :

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$\sf \implies {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}}$

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$\textsf{☣ \;Putting\; Value \;of\; π \;=\; 180°}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀_________________________________________________________________________

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$\sf \implies {sin^2\ \dfrac{180^{\circ}}{6}\ +\ cos^2\ \dfrac{180^{\circ}}{3}\ -\ tan^2\ \dfrac{180^{\circ}}{6}}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

$\sf \implies {sin^2\ 30^{\circ}\ +\ cos^2\ 60^{\circ}\ -\ tan^2\ 45^{\circ}}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;$

$\sf \implies{(sin\ 30^{\circ})^2\ +\ (cos\ 60^{\circ})^2\ -\ (tan\ 45^{\circ})^2}$

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$\textsf{☣\; Now \;putting \;values\; required\; :}$

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• ➽ $\sf {sin\ 30^{\circ}\ =\ 1/2}$

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• ➽ $\sf {cos\ 60^{\circ}\ =\ 1/2}$

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• ➽ $\sf {tan\ 45^{\circ}\ =\ 1}$

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$\sf \implies {\bigg(\dfrac{1}{2} \bigg)^2\ +\ \bigg(\dfrac{1}{2} \bigg)^2\ -\ 1^2}$

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$\sf \implies{\dfrac{1}{4}\ +\ \dfrac{1}{4}\ -\ 1}$

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$\sf \implies{\dfrac{1\ +\ 1\ -\ 4}{4}}$

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$\sf \implies{\dfrac{2\ -\ 4}{4}}$

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$\sf \implies {\dfrac{-2}{4}}$

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$\sf \implies{\dfrac{-1}{2}}$

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$\sf \implies {R.H.S}$

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• $\sf {sin^2\ \dfrac{\pi}{6}\ +\ cos^2\ \dfrac{\pi}{3}\ -\ tan^2\ \dfrac{\pi}{4}\ =\ - \dfrac{1}{2}}$

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$\green\bigstar$$\textsf{Hence\ Proved !}$$\pink\bigstar$