Question

prove that (sinA - sec A )^2 +(cos A -cosec A )^2 = (1-sec A * cosec A)^2​

Answer 1

❥︎ Question :-

Prove that :-

$\bf \:(sinA - sec A )^2 +(cos A -cosec A )^2 = (1-sec A * cosec A)^2$

❥︎ Solution :-

Consider

$\bf \:(sinA - sec A )^2 +(cos A -cosec A )^2$

$\bf\implies \:(sinA - \dfrac{1}{cos A} )^2 +(cos A -\dfrac{1}{sinA} )^2 =$

$\bf\implies \: {(\dfrac{sinAcos A - 1}{cos A}) }^{2} + {(\dfrac{sinAcos A - 1}{sinA}) }^{2}$

$\bf\implies \:\dfrac{ {(sinAcos A - 1)}^{2} }{ {(cos A)}^{2} } + \dfrac{ {(sinAcos A - 1)}^{2} }{ {(sinA)}^{2} }$

$\bf\implies \: {(sinAcos A - 1)}^{2}( \dfrac{1}{ {cos }^{2} A} + \dfrac{1}{ {sin}^{2}A } )$

$\bf\implies \: {(sinAcos A - 1)}^{2} (\dfrac{ {sin}^{2}A + {cos}^{2} A }{ {sin}^{2}A {cos}^{2} A } )$

$\bf\implies \: {(sinAcos A - 1)}^{2} \times \dfrac{1}{ {sin}^{2}A {cos}^{2}A }$

$\bf\implies \: {(\dfrac{sinAcos A - 1}{sinAcos A} )}^{2}$

$\bf\implies \: {(\dfrac{sinAcos A}{sinAcos A} - \dfrac{1}{sinAcos A} )}^{2}$

$\bf\implies \: {(1 - secA \: cosecA)}^{2}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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Additional Information:-

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

❥︎Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

❥︎Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

❥︎Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\bf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}TrigonometryTable∠AsinAcosAtanAcosecAsecAcotA0∘010∞1∞30∘212331232345∘2121122160∘232133223190∘10∞1∞0$