Question

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Answer 1

Step-by-step explanation:

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Answer 2

❥︎Question :-

$\bf \:If \: \sqrt{2} sinθ = 1, \: find \: {sec}^{2} θ - {cosec}^{2} θ$

❥︎Answer

❥︎Given :-

$\bf \:\sqrt{2} sinθ = 1$

❥︎To Find :-

$\bf \:{sec}^{2} θ - {cosec}^{2} θ$

❥︎Solution :-

$\bf \:\sqrt{2} sinθ = 1$

$\bf\implies \:sinθ = \dfrac{1}{ \sqrt{2} }$

$\bf\implies \:sinθ = sin45°$

$\bf\implies \:θ = 45°$

❥︎Now, Consider

$\bf \:{sec}^{2} θ - {cosec}^{2} θ$

❥︎On substituting the value of θ = 45°, we get

$\bf\implies \:{sec}^{2} 45° - {cosec}^{2} 45°$

$\bf\implies \: {( \sqrt{2} )}^{2} - {( \sqrt{2}) }^{2}$

$\bf\implies \:2 - 2 = 0$

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❥︎Additional Information:-

❥︎Relationship between sides and T ratios

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

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