Question

write all formula of trigonometry and represent the trigonometric ratios of angle ​

Answer 1

$\bigstar { \pmb{ \red { \sf \: trigonometric \: ratios }}}$

$\implies \sf \pink{ \sin \theta = \dfrac{opposite \: side}{hypotenuse} }$

$\implies \sf \green{ \cos \theta = \dfrac{adjacent \: side}{hypotenuse} }$

$\implies \sf \pink{ \tan \theta = \dfrac{opposite \: side}{adjacent \: side} }$

$\implies \sf \green{ \cosec \theta = \dfrac{hypotenuse}{opposite\: side} = \dfrac{1}{ \sin \theta} }$

$\implies \sf \pink{ \sec\theta = \dfrac{hypotenuse}{adjacent \: side} = \dfrac{1}{ \cos \theta } }$

$\implies \sf \green{ \cot \theta = \dfrac{adjacent \: side}{opposite \: side} = \dfrac{1}{ \tan \theta} }$

$\bigstar { \pmb{ \red { \sf \: trigonometric \: identities }}}$

$\longrightarrow\sf \orange{ { \sin}^{2} \theta + { \cos }^{2} \theta = 1 }$

$\longrightarrow\sf \blue{ { \sec}^{2} \theta - { \ tan}^{2} \theta = 1 }$

$\longrightarrow\sf \orange{ { \cosec}^{2} \theta + { \cot }^{2} \theta = 1 }$

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

$\bigstar { \pmb{ \red { \sf \: Ratios\:of\:complementary\:angles }}}$

• sin∅ = cos(90-∅)
• cos∅ = sin(90-∅)
• tan∅ = cot(90-∅)
• cot∅ = tan(90-∅)
• sec∅ = cosec(90-∅)
• cosec∅ = sec(90-∅)

_______________________________

hope it's beneficial :D

Answer 2

${\Huge{ \red{\underline{\underline{\bf{\maltese \pink{ Ànswer: \: }}}}}}}$

$\huge\bigstar{ \pmb{ \red { \sf \: trigonometric \: ratios }}}★$

$\implies\sf\pink{ \sin \theta = \dfrac{opposite \: side}{hypotenuse} }$

$\implies \sf \red{ \cos \theta = \dfrac{adjacent \: side}{hypotenuse} }$

$\implies \sf \green{ \tan \theta = \dfrac{opposite \: side}{adjacent \: side} }$

$\implies\sf \blue{ \cosec \theta = \dfrac{hypotenuse}{opposite\: side} = \dfrac{1}{ \sin \theta} }$

$\implies \sf \red{ \sec\theta = \dfrac{hypotenuse}{adjacent \: side} = \dfrac{1}{ \cos \theta } }$

$\implies \sf \orange{ \cot \theta = \dfrac{adjacent \: side}{opposite \: side} = \dfrac{1}{ \tan \theta} }$

$\huge\bigstar { \pmb{ \red { \sf \: trigonometric \: identities }}}★$

$\boxed \longrightarrow\sf\orange{ { \sin}^{2} \theta + { \cos }^{2} \theta = 1 }$

$\longrightarrow\sf \blue{ { \sec}^{2} \theta - { \ tan}^{2} \theta = 1 }$

$\longrightarrow\sf \red{ { \cosec}^{2} \theta + { \cot }^{2} \theta = 1 }$

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

$\huge\bigstar { \pmb{ \pink { \sf \: Ratios\:of\:complementary\:angles }}}★$

sin∅ = cos(90-∅)

cos∅ = sin(90-∅)

tan∅ = cot(90-∅)

cot∅ = tan(90-∅)

sec∅ = cosec(90-∅)

cosec∅ = sec(90-∅)

--------------------×××××--------------------