Question

Cot^4[30]-sin^2[60]+sec^2[45]

Answer 1

Solution:

cot⁴30° - sin²60° + sec²45°

We know that

• cot30° = √3
• sin60° = √3/2
• sec45° = √2

Substituting the values we have

→ ( √3 )⁴ - [√3/2]² + ( √2 )²

→ [(√3)²]² - ( √3 )²/2² + 2

→ 3² - 3/4 + 2

→ 11 - ¾

→ 44 - 3/4

→ cot⁴30° - sin²60° + sec²45° = 41/4

Hence, cot⁴30° - sin²60° + sec²45° = 41/4

__________________________

Knowledge enhancer:

Identities:

♦ sin²A + cos²A = 1

♦ sec²A - tan²A = 1

♦ cosec²A - cot²A = 1

Formulas:

♦ sin(A + B) = sinA.cosB + cosA.sinB

♦ sin(A - B) = sinA.cosB - cosA.sinB

♦ cos(A - B) = cosA.cosB + sinA.sinB

♦ cos(A + B) = cosA.cosB - sinA.sinB

♦ tan(A + B) = sin(A + B)/cos(A + B)

=> sinAcosB + cosA.sinB/cosA.cosB - sinA.sinB

Answer 2

$\red{\huge{\boxed{\sf{Question:}}}}$

$\bf Cot^4 30^{o} - sin^2 60^{o} + sec^2 45^{o}$ = ?

$\red{\huge{\boxed{\sf{Solution:}}}}$

Let's look at some values

$\begin{array}{| c | c | c | c | c | c |}\cline{1-6} \bf Angles & \bf 0^{o} & \bf 30^{o} & \bf 45^{o} & \bf 60^{o} & \bf 90^{o} \\\cline{1-6} \tt Sin \theta & 0 & \dfrac{1}{2} & \dfrac{1}{\sqrt{2}}& \dfrac{\sqrt{3}}{2}& 1\\\cline{1-6} \tt cos \theta & 1 & \dfrac{\sqrt{3}}{2} &\dfrac{1}{\sqrt{2}}&\dfrac{1}{2}&0\\\cline{1-6} \tt tan \theta & 0 & \dfrac{1}{\sqrt{3}} & 1& \sqrt{3} & \infty \\\cline{1-6} \tt cosec \theta & \infty & 2 & \sqrt{2} & \dfrac{2}{\sqrt{3}} &1\\\cline{1-6} \tt sec \theta & 1 & \dfrac{2}{\sqrt{3}} & \sqrt{2} & 2 & \infty \\\cline{1-6} \tt cot \theta & \infty & \sqrt{3} &1& \dfrac{1}{\sqrt{3}} &0\\\cline{1-6}\end{array}$

Here, '$\red{\infty}$' stands for undefined(not defined) values.

Putting the value of each

$\bf = (\sqrt{3})^{4} - (\frac{\sqrt{3}}{2})^{2} + (\sqrt{2})^{2}$

$\bf = (3)^{2} - \frac{(\sqrt{{3}})^{2}}{(2)^{2}} +2$

$\bf = 9 - \frac{3}{4} +2$

$\bf = 9 +2 - \frac{3}{4}$

$\bf = 11 - \frac{3}{4}$

Taking LCM 4

$\bf = \frac{44-3}{4}$

$\bf = \frac{41}{4}$

Hence

The value of

$\bf Cot^4 30^{o} - sin^2 60^{o} + sec^2 45^{o} = \boxed{\boxed{\frac{41}{4}}}$