Question

2 tan-45° + cos²30° - sin’609​

Answer 1

Answer:

${\large{\pmb{\underline{\sf{★Correct \: Question...}}}}}$

➞ 2tan²45° + cos²30° - sin²60°

${\large{\pmb{\underline{\sf{★Solution...}}}}}$

Using Trigonometry Table,

• Tan45° = 1

• Cos30° = √3/2

• Sin60° = √3/2

By substituting values,

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

${\large{\pmb{\underline{\sf{★Learn \: More...}}}}}$

Learn Trigonometry Table here,

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