Question

Evaluate ( cos²45°•sec²30°) (sin²30°+4cot²45°- sec²60°)​

Answer 1

Answer:

• 2/3.

Step-by-step explanation:

We have,

⇒ (cosec²45° sec²30°) (sin²30° + 4cot²45° - sec²60°)

We know,

⇒ cosec 45° = √2

⇒ sec 30° = 2/√3

⇒ sin 30° = 1/2

⇒ cot 45° = 1

⇒ sec 60° = 1

Substituting,

⇒ [(√2)².[2/√3]²].[(1/2) + 4(1)² . (2)²]

⇒ [2(4/3)] [1/4 + 4 - 4]

⇒ [3(4/3)(1/4)]

⇒ 2/3

∴ 2/3 is the value of the expression.

Learn more:

⇒ sin 0° = 0  

⇒ sin 30° = 1/2  

⇒ sin 45° = 1/√2  

⇒ sin 60° = √3/2  

⇒ sin 90° = 1

=> cos Ф is the reverse of sin Ф

⇒ cos 0° = 1  

⇒ cos 30° = √3/2  

⇒ cos 45° = 1/√2  

⇒ cos 60° = 1/2  

⇒ cos 90° = 0

=> tan Ф = sin Ф/cos Ф

⇒ tan 0° = sin 0°/cos 0°  

⇒ tan 0° = 1/0  

⇒ tan 0 ° = 1

⇒ tan 30° = sin 30°/cos 30°  

⇒ tan 30° = (1/2)/(√3/2)  

⇒ tan 30° = 1/√3

⇒ tan 45° = sin 45°/cos°  

⇒ tan 45° = (1/√2)/(1/√2)  

⇒ tan 45° = 1

⇒ tan 60° = sin 60°/cos 60°  

⇒ tan 60° = (√3/2)/((1/2)  

⇒ tan 60° = √3

⇒ tan 90° = sin 90°/cos 90°  

⇒ tan 90° = 1/0  

⇒ tan 90° = undefined

Others:

=> tan Ф = 1/cot Ф  

=> sin Ф = 1/cosec Ф  

=> sec Ф = 1/cos Ф

Answer 2

$\large\underline{\sf{Solution-}}$

Given Trigonometric expression is

$\rm :\longmapsto\: [{cos}^{2}45\degree {sec}^{2}30\degree ]\bigg[ {sin}^{2}30\degree + {4cot}^{2}45\degree - {sec}^{2}60\degree \bigg]$

We know, Trigonometric table of standard angles,

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

We have,

$\boxed{ \tt{ \: cos45\degree = \frac{1}{ \sqrt{2} }}}$

$\boxed{ \tt{ \: sec30\degree = \frac{2}{ \sqrt{3} }}}$

$\boxed{ \tt{ \: sin30\degree = \frac{1}{2}}}$

$\boxed{ \tt{ \: cot45\degree = 1}}$

$\boxed{ \tt{ \: sec60\degree = 2}}$

So, on substituting the values in

$\rm :\longmapsto\: [{cos}^{2}45\degree {sec}^{2}30\degree ]\bigg[ {sin}^{2}30\degree + {4cot}^{2}45\degree - {sec}^{2}60\degree \bigg]$

we get

$\rm \: = \: {\bigg[\dfrac{1}{ \sqrt{2} } \bigg]}^{2} {\bigg[\dfrac{2}{ \sqrt{3} } \bigg]}^{2} \bigg( {\bigg[\dfrac{1}{2} \bigg]}^{2} + 4 {(1)}^{2} - {(2)}^{2} \bigg)$

$\rm \: = \:\dfrac{1}{2} \times \dfrac{4}{3} \times \bigg[\dfrac{1}{4} + 4 - 4\bigg]$

$\rm \: = \:\dfrac{1}{2} \times \dfrac{4}{3} \times \dfrac{1}{4}$

$\rm \: = \:\dfrac{1}{6}$