Question

prove that: 2 sin square pi/6+ cosec square 7 pi/6×cos square pi/3=3/2​

Answer 1

Step-by-step explanation:

pls check the one by one all three attachments for all the step.

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Answer 2

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Prove that :-

• $\implies \bf\rm2 { \sin }^{2} \dfrac{ \pi}{6} + { \csc }^{2} \dfrac{7 \pi}{6} \times { \cos} ^{2} \dfrac{ \pi}{3}$

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Taking LHS :-

• $= 2 { \sin }^{2} \dfrac{ \pi}{6} + { \csc }^{2} \dfrac{7 \pi}{6} \times { \cos}^{2} \dfrac{ \pi}{3}$
• $\\ \\ = 2 { \sin}^{2} \frac{ \pi}{6} + { \csc }^{2} \bigg( \pi + \frac{ \pi}{6} \bigg) \times { \cos }^{2} \frac{ \pi}{3}$
• $\\ \\ = 2 {\bigg( \frac{1}{2} \bigg)}^{2} - { \csc }^{2} \bigg( \frac{ \pi}{6} \bigg) \times { \ \cos }^{2} \frac{ \pi}{3}$
• $\\ \\ = \cancel2 \times \frac{ \cancel1}{ \cancel4} + {( - 2)}^{2} \times{\bigg( \frac{1}{2} \bigg)}^{2}$
• $\\ \\ = \frac{1}{2} + \frac{ \cancel4}{1} \times \frac{1}{ \cancel4}$
• $\\ \\ = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{1 + 2}{2} = \dfrac{3}{2}$

LHS = RHS

$\huge{ \sf \underline{{ ╬Hence \:Proved ╬ \: \: }}}$