sin25°/sec65° + cos25°/cosec65° + tan25°/cot65°

Question:

$\dfrac{\sin(25^\circ)}{\sec(65^\circ)} + \dfrac{\cos(25^\circ)}{\text{cosec}(65^\circ)} + \dfrac{\tan(25^\circ)}{\cot(65^\circ)}$

$\dfrac{\sin(25^\circ)}{\sec(65^\circ)} + \dfrac{\cos(25^\circ)}{\text{cosec}(65^\circ)} + \dfrac{\tan(25^\circ)}{\cot(65^\circ)}$

$\Rightarrow \dfrac{\sin(90-65)^\circ}{\sec 65^\circ} + \dfrac{\cos(90-65)^\circ}{\text{cosec}65^\circ} + \dfrac{\tan(90-65)^\circ}{\cot 65^\circ}$

$\Rightarrow \dfrac{\cos65^\circ}{\sec 65^\circ} + \dfrac{\sin65^\circ}{\text{cosec}65^\circ} + \dfrac{\cot65^\circ}{\cot 65^\circ}$

$\Rightarrow (cos65^\circ\cdot cos65^\circ) + (sin65^\circ\cdot sin65^\circ) + 1$

$\Rightarrow cos^{2}65^\circ + sin^{2}65^\circ + 1$

$\Rightarrow 1 + 1 \qquad [\because sin^2\theta + cos^2\theta = 1]$

$\Rightarrow 2 \quad(Answer)$

Formulas Used:

$sin^2\theta + cos^2\theta = 1$
$\sin(90-\theta) = \cos \theta$
$\cos(90-\theta) = \sin \theta$
$\tan(90-\theta) = \cot \theta$