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Sagot :
Answer:
See below ↓
Step-by-step explanation:
We need to prove :
⇒ cos⁻¹ [tex]\frac{12}{13}[/tex] + sin⁻¹ [tex]\frac{3}{5}[/tex] = tan⁻¹ [tex]\frac{56}{65}[/tex]
Let's simplify the LHS.
- cos⁻¹ [tex]\frac{12}{13}[/tex] + sin⁻¹ [tex]\frac{3}{5}[/tex]
Convert the inverse cos and sin functions into inverse tan functions
- tan⁻¹ [tex]\frac{5}{12}[/tex] + tan⁻¹ [tex]\frac{3}{4}[/tex]
- [∴This can be found taking a right triangle and labeling the sides, and then using Pythagorean Theorem, we can find the missing side and take the ratio of tan]
Identity
- tan⁻¹ x + tan⁻¹ y = tan⁻¹ [tex]\frac{x+y}{1-xy}[/tex]
Using this identity, we can simplify our earlier equation!
⇒ tan⁻¹ [(5/12 + 3/4)/(1 - (5/12 x 3/4))]
⇒ tan⁻¹ [(20 + 36) / (48 - 15)
⇒ tan⁻¹ (56/65)
⇒ RHS
⇒ Proved ∴√
[tex]\text{L.H.S}\\\\=\cos^{-1} \dfrac{12}{13} + \sin^{-1} \dfrac 35\\\\=\sin^{-1} \dfrac 5{13} + \sin^{-1} \dfrac 35\\\\[/tex]
[tex]=\sin^{-1}\left[\dfrac 5{13}\sqrt{1- \left(\dfrac 35 \right)^2} + \dfrac 35\sqrt{1-\left(\dfrac 5{13} \right)^2} \right]\\\\=\sin^{-1} \left(\dfrac 5{13} \sqrt{1-\dfrac 9{25} }+\dfrac 35 \sqrt{1-\dfrac{25}{169}} \right)\\\\=\sin^{-1} \left(\dfrac 5{13} \sqrt{\dfrac{16}{25}}+\dfrac 35 \sqrt{\dfrac{144}{169}} \right)\\\\=\sin^{-1} \left(\dfrac{5}{13} \cdot \dfrac 45 + \dfrac 35 \cdot \dfrac{12}{13} \right)\\[/tex]
[tex]=\sin^{-1} \left(\dfrac 4{13} +\dfrac{36}{65}\right)\\\\=\sin^{-1} \left(\dfrac{20}{65} + \dfrac{36}{65} \right)\\\\=\sin^{-1} \left(\dfrac{20+36}{65} \right)\\\\=\sin^{-1} \left(\dfrac{56}{65} \right)\\\\=\text{R.H.S}[/tex]
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