arc tan (x) + arc cot ​​(x) = π/2

$arctan(x)+arccot(y)=\frac{\pi }{2}$

Kita akan belajar bagaimana membuktikan sifat dari invers fungsi trigonometri

Arc tan (x) + arc cot ​​(x) = $\frac{1}{2}$π (yaitu, tan⁻¹ x + cot⁻¹ x = $\frac{1}{2}$π).

Bukti: misalkan, tan⁻¹ x = θ

Oleh karena itu, x = tan θ

x = cot ($\frac{1}{2}$π – θ), [Karena, cot ($\frac{1}{2}$π – θ) = tan θ]

⇒ cot⁻¹ x = $\frac{1}{2}$π – θ

⇒ cot⁻¹ x = $\frac{1}{2}$π – tan⁻¹ x, [Karena, θ = tan⁻¹ x]

⇒ cot⁻¹ x + tan⁻¹ x = $\frac{1}{2}$π

⇒ tan⁻¹ x + cot⁻¹ x = $\frac{1}{2}$π

Oleh karena itu, tan⁻¹ x + cot⁻¹ x = $\frac{1}{2}$π.

Contoh Soal

Dengan tan⁻¹ x + cot⁻¹ x = $\frac{1}{2}$π,

buktikan bahwa, tan⁻¹ $\frac{4}{3}$ + tan⁻¹ $\frac{12}{5}$ = π – tan⁻¹ $\frac{56}{33}$.

Jawab:

Kita tahu bahwa tan⁻¹ x + cot⁻¹ x = $\frac{1}{2}$π,

⇒ tan⁻¹ x = $\frac{1}{2}$π – cot⁻¹ x

⇒ tan⁻¹ $\frac{4}{3}$ = $\frac{1}{2}$π – cot⁻¹$\frac{4}{3}$

dan

tan⁻¹ $\frac{12}{5}$ = $\frac{1}{2}$π – cot⁻¹ $\frac{12}{5}$

untuk tan⁻¹ $\frac{4}{3}$ + tan⁻¹ $\frac{12}{5}$

= $\frac{1}{2}$π – cot⁻¹ $\frac{4}{3}$ + $\frac{1}{2}$π – cot⁻¹ $\frac{12}{5}$

[karena, tan⁻¹ $\frac{4}{3}$ = $\frac{1}{2}$π – cot⁻¹ $\frac{4}{3}$ dan tan⁻¹ $\frac{12}{5}$ = $\frac{1}{2}$π – cot⁻¹ $\frac{12}{5}$]

= π – [cot⁻¹ $\frac{4}{3}$ + cot⁻¹ $\frac{12}{5}$]

= π – [tan⁻¹ $\frac{3}{4}$ + tan⁻¹ $\frac{5}{12}$]

= π – tan⁻¹$\left ( \frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}.\frac{5}{12}} \right )$

= π – tan⁻¹($\frac{14}{12}$ x $\frac{48}{33}$)

= π – tan⁻¹ $\frac{56}{33}$

Invers Fungsi Trigonometri

• Nilai Umum dan Pokok sin⁻¹ x
• Nilai Umum dan Pokok dari cos⁻¹ x
• Nilai Umum dan Pokok dari tan⁻¹ x
• Nilai Umum dan Pokok dari csc⁻¹ x
• Nilai Umum dan Pokok dari sec⁻¹ x
• Nilai Umum dan Pokok dari cot⁻¹ x
• Nilai Pokok Invers Fungsi Trigonometri  
• Nilai Umum Invers Fungsi Trigonometri  
• arcsin(x) + arccos(x) = 𝜋/2
• arctan(x) + arccot(x) = 𝜋/2
• $tan^{-1}x+tan^{-1}y=tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
• $tan^{-1}x-tan^{-1}y=tan^{-1}\left ( \frac{x-y}{1+xy} \right ) $
• $tan^{-1}x+tan^{-1}y+tan^{-1}z=tan^{-1}\left ( \frac{x+y+z-xyz}{1-xy-yz-xz} \right ) $
• $cot^{-1}x+cot^{-1}y=cot^{-1}\left (\frac{xy-1}{y+x} \right ) $
• $cot^{-1}x-cot^{-1}y=cot^{-1}\left (\frac{xy+1}{y-x} \right ) $
• $sin^{-1}x+sin^{-1}y=sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$
• $sin^{-1}x-sin^{-1}y=sin^{-1}(x\sqrt{1-y^2}-y\sqrt{1-x^2})$
• $cos^{-1}x+cos^{-1}y=cos^{-1}(xy-\sqrt{1-x^2} \sqrt{1-y^2})$
• $cos^{-1}x-cos^{-1}y=cos^{-1}(xy+\sqrt{1-x^2} \sqrt{1-y^2})$
• $2sin^{-1}x=sin^{-1}(2x\sqrt{1-x^2})$
• $2cos^{-1}x=cos^{-1}(2x^2-1)$
• $2tan^{-1}x=tan^{-1}\left ( \frac{2x}{1-x^2} \right )=sin^{-1}\left ( \frac{2x}{1+x^2} \right )=cos^{-1}\left ( \frac{1-x^2}{1+x^2} \right )$
• $3sin^{-1}x=sin^{-1}(3x-4x^3)$
• $3cos^{-1}x=cos^{-1}(4x^3-3x)$
• $3tan^{-1}x=tan^{-1}\left ( \frac{3x-x^3}{1-3x^2} \right )$
• Kumpulan Rumus Invers Fungsi Trigonometri  
• Nilai Utama Invers Fungsi Trigonometri
• Soal-soal & Pembahasan Invers Fungsi Trigonometri