Question

If √2 sin Ø - 1, find the value of sec²Ø - cosec²Ø​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\underline{\underline{\sf{\maltese\:\:Given}}}$

• √2sinθ - 1

$\underline{\underline{\sf{\maltese\:\:To\: Find}}}$

• sec²θ - cosec²θ

$\underline{\underline{\sf{\maltese\:Calculations \:}}}$

STEP 1 :-

⠀⠀⠀⠀⠀⠀As given √2sinθ - 1

Therefore,

$\longrightarrow \sinθ = \frac{1 } { \sqrt{2} }$

$\longrightarrow \sinθ = 45°$

STEP 2 :-

⠀⠀⠀⠀⠀⠀θ = 45°

STEP 3 :-

⠀⠀⠀⠀⠀⠀Now, let's find sec²θ - cosec²θ

substituting the value of θ, we get :-

$\longrightarrow ( { \sec}^{2} 45°)^{2} - ( { \cosec}^{2} 45°)^{2}$

Putting the value of sec²45° and cosec²45°, we get :-

$\longrightarrow( \sqrt{2} )^{2} - ( \sqrt{2} )^{2}$

$\longrightarrow2 - 2$

$\longrightarrow0$

Hence, the value of sec²θ - cosec²θ is 0

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