Question

Please help me with this question

Answer 1

Answer:

${ \large{ \pmb{ \sf{Given... }}}}$

• A = 60°

• B = 30°

$\large{ \pmb{ \sf{To \: Verify... }}}$

Cos(A - B) = CosA CosB + SinA SinB

$\: { \large{ \pmb{ \sf{Solution... }}}}$

By Substituting A and B values,

${ \implies{ \sf{cos(60° - 30°) = cos60 °\times cos30° + sin60 °\times sin30}}}$

${ \implies{ \sf{cos30°= \frac{1}{2} \times \frac{ \sqrt{3} }{2} + \frac{ \sqrt{3} }{2} \times \frac{1}{2} }}}$

${ \implies{ \sf{ \frac{ \sqrt{3} }{2} = \frac{ \sqrt{3} }{4} + \frac{ \sqrt{3} }{4} }}}$

${ \implies{ \sf{ \frac{ \sqrt{3} }{2} = \frac{ \sqrt{3} + \sqrt{3} }{4} }}}$

$\: { \implies{ \sf{ \frac{ \sqrt{3} }{2} = \frac{2 \sqrt{3} }{4} }}}$

$\: \: { \implies{ \sf{ \frac{ \sqrt{3} }{2} = \frac{{ \cancel{2 }}\sqrt{3} }{2 \times { \cancel{2}}} }}}$

$\: \: { \implies{ \sf{ \frac{ \sqrt{3} }{2} = \frac{\sqrt{3} }{2} }}}$

Hence Proved

${ \large{ \pmb{ \sf{More \: Information...}}}}$

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

Answer 2

Answer:

tooo long answer given by anybody

how can he/she write this much

Step-by-step explanation:

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