Question

prove that 2cos²π/3 + 3/4sec²π/4 + 4sin²π/6 = cot² π/6

​

Answer 1

Step-by-step explanation:

$\sf \underline \red{Answer:-}$

Here,

• cos π/3 = 1/2
• sec π/4 = √2
• sin π/6 = 1/2
• cot π/6 = √3

LHS:-

$\sf \longmapsto \: 2 {cos}^{2} \frac{\pi}{3} + \frac{3}{4} {sec}^{2} \frac{\pi}{4} + 4 {sin}^{2} \frac{\pi}{6}$

$\sf \longmapsto \:2 {( \frac{1}{2} })^{2} + \frac{3}{4} {( \sqrt{2} })^{2} + 4 {( \frac{1}{2} )}^{2}$

$\sf \longmapsto \: \frac{1}{2} + \frac{3}{2} + 1$

$\sf \longmapsto \:3$

RHS:-

$\sf \longmapsto \: {cot}^{2} \frac{\pi}{6}$

$\sf \longmapsto \: {( \sqrt{3} })^{2}$

$\sf \longmapsto \:3$

•°• LHS = RHS

Answer 2

Step-by-step explanation:

Step-by-step explanation:

\sf \underline \red{Answer:-}

Answer:−

Here,

cos π/3 = 1/2

sec π/4 = √2

sin π/6 = 1/2

cot π/6 = √3

LHS:-

\begin{gathered} \sf \longmapsto \: 2 {cos}^{2} \frac{\pi}{3} + \frac{3}{4} {sec}^{2} \frac{\pi}{4} + 4 {sin}^{2} \frac{\pi}{6} \\ \end{gathered}

⟼2cos

2

3

π

+

4

3

sec

2

4

π

+4sin

2

6

π

\begin{gathered}\sf \longmapsto \:2 {( \frac{1}{2} })^{2} + \frac{3}{4} {( \sqrt{2} })^{2} + 4 {( \frac{1}{2} )}^{2} \\ \end{gathered}

⟼2(

2

1

)

2

+

4

3

(

2

)

2

+4(

2

1

)

2

\begin{gathered}\sf \longmapsto \: \frac{1}{2} + \frac{3}{2} + 1 \\ \end{gathered}

⟼

2

1

+

2

3

+1

\sf \longmapsto \:3⟼3

RHS:-

\sf \longmapsto \: {cot}^{2} \frac{\pi}{6}⟼cot

2

6

π

\sf \longmapsto \: {( \sqrt{3} })^{2}⟼(

3

)

2

\sf \longmapsto \:3⟼3

•°• LHS = RHS