Question

Prove that :-
→ (1-cos²@)sec²@ = 1
[@ = theta]​

Answer 1

Given

• (1 - cos²θ) sec²θ = 1

To prove

• L.H.S = R.H.S

Solution

$\large\tt\longmapsto{\color{cyan}{Taking\: L.H.S}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {L.H.S = (1 - cos^2θ) sec^2θ}$

⠀

$\sf\pink{⟶}$ 1 - sin²θ = cos²θ

$\tt:\implies\: \: \: \: \: \: \: \: {L.H.S = cos^2θ.sec^2θ}$

⠀

$\huge{[}$ $\because{secθ = \dfrac{1}{cosθ}{\therefore{sec^2θ = \dfrac{1}{cos^2θ}}}}$ $\huge{]}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {L.H.S = cos^2θ (\dfrac{1}{cos^2θ})}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {L.H.S = \cancel{cos^2θ} (\dfrac{1}{\cancel{cos^2θ}})}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {L.H.S = 1 = R.H.S}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {L.H.S = R.H.S}$

$\mathcal\color{lime}{Hence\: proved}$

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$\: \: \: \: \: \: \: \: \: \: \: \: \boxed{\tt{\bigstar{MORE\: TO\: KNOW{\bigstar}}}}$

⠀

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

⠀

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Answer 2

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Gɪᴠᴇɴ -

• (1 - ᴄᴏꜱ²Θ) ꜱᴇᴄ²Θ = 1

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Tᴏ Pʀᴏᴠᴇ -

• ʟ.ʜ.ꜱ = ʀ.ʜ.ꜱ

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Sᴏʟᴜᴛɪᴏɴ -

$\large\bf\longmapsto{\color{red}{Taking\: L.H.S}}$⠀

ㅤㅤ

$\tt:\implies{L.H.S = (1 - cos^2θ) sec^2θ}$

ㅤㅤ

$\tt:\implies {L.H.S = cos^2θ.sec^2θ}$

ㅤㅤ

${\huge[}\because{secθ = \dfrac{1}{cosθ}{\therefore{sec^2θ = \dfrac{1}{cos^2θ}}}}\huge{]}$

ㅤㅤ

$\tt:\implies{L.H.S = cos^2θ (\dfrac{1}{cos^2θ})}$

ㅤㅤ

$\tt:\implies{L.H.S = \cancel{cos^2θ} (\dfrac{1}{\cancel{cos^2θ}})}$

ㅤㅤ

$\tt:\implies {L.H.S = 1 = R.H.S}$

ㅤㅤ

$\tt:\implies {L.H.S = R.H.S}$

ㅤㅤ

$\huge\tt\color{red}{Hence\: proved}$

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