Model 1: Problems based on proving trigonometric identities based on 2A. A. (Prove that): 1/sin 10° cos 10° = √3 OR cosec 10° - √3 sec 10° = 4

$R = \sqrt{a^2 + b^2} = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = \sqrt{4} = 2$
अंशलाई 2 ले गुणन र भाग गरौं:

$$\text{Numerator} = 2 \left( \frac{1}{2} \cos 10^\circ - \frac{\sqrt{3}}{2} \sin 10^\circ \right)$$

हामीलाई थाहा छ $\cos 60^\circ = \frac{1}{2}$ र $\sin 60^\circ = \frac{\sqrt{3}}{2}$।

$$\text{Numerator} = 2 (\cos 60^\circ \cos 10^\circ - \sin 60^\circ \sin 10^\circ)$$

$\cos(A+B) = \cos A \cos B - \sin A \sin B$ सूत्र प्रयोग गर्दा:

$$\text{Numerator} = 2 \cos(60^\circ + 10^\circ) = 2 \cos 70^\circ$$

$\sin 2A = 2 \sin A \cos A \implies \sin A \cos A = \frac{\sin 2A}{2}$

$$\text{Denominator} = \sin 10^\circ \cos 10^\circ = \frac{\sin(2 \times 10^\circ)}{2} = \frac{\sin 20^\circ}{2}$$

$$\text{LHS} = \frac{2 \cos 70^\circ}{\frac{\sin 20^\circ}{2}}$$ $$\text{LHS} = \frac{4 \cos 70^\circ}{\sin 20^\circ}$$

हामीलाई थाहा छ $\cos(90^\circ - \theta) = \sin \theta$।
त्यसैले, $\cos 70^\circ = \cos(90^\circ - 20^\circ) = \sin 20^\circ$।

$$\text{LHS} = \frac{4 \sin 20^\circ}{\sin 20^\circ}$$ $$\text{LHS} = 4$$

यो दायाँ पक्ष (RHS) सँग बराबर छ।

$$\text{LHS} = \text{RHS}$$

अतः, प्रमाणित भयो।

$\text{प्रमाणित भयो}$