Question

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All 10th class maths formula.
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Answer 1

Answer:

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

Answer 2

(a + b)2= a2 + 2ab + b2

(a – b)2 = a2– 2ab + b2

a2– b2 = (a + b) (a – b)

(a + b)3 = a3+ b3 + 3ab(a + b)

(a – b)3 = a3– b3 – 3ab(a – b)

a3+ b3 = (a + b) (a2 – ab + b2)

a3– b3 = (a – b) (a2 + ab + b2)

a4– b4 = (a2)2 – (b2)2 = (a2 + b2) (a2 – b2) = (a2 + b2) (a + b) (a – b)

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac

(a + b – c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ca

(a – b + c)2 = a2+ b2 + c2 – 2ab – 2bc + 2ca

(a – b – c)2 = a2+ b2 + c2 – 2ab + 2bc – 2ca

a3+ b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)