SATHEE: Shortcut Methods

JEE Main Numerical:

$\sin 39 \degree = \frac{\sqrt{10} - \sqrt{2}}{4}$
$\cos 23 \degree = \frac{\sqrt{2} + \sqrt{2}}{2 \sqrt{2}} = \frac{\sqrt{2}}{2}$
$t a n 15 \degree = \frac{\sqrt{3} - \sqrt{3}}{2 \sqrt{3}} = \frac{1}{\sqrt{3}}$
$s i n^{2} 45 \degree + c o s^{2} 45 \degree = {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$s e c 60 \degree - c o s e c 60 \degree = 2 - \frac{2 \sqrt{3}}{3} = \frac{2 (3 - \sqrt{3})}{3}$
$c o t 45 \degree = \frac{c o s 45 \degree}{s i n 45 \degree} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$
$t a n 60 \degree + t a n 30 \degree = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} (3 + 1)}{\sqrt{3}} = \frac{4 \sqrt{3}}{3}$
$s i n 30 \degree + s i n 60 \degree + s i n 90 \degree = \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 = \frac{1 + 2 \sqrt{3} + 2}{2} = \frac{3 + \sqrt{3}}{2}$
$c o s 30 \degree + c o s 60 \degree + c o s 90 \degree = \frac{\sqrt{3}}{2} + \frac{1}{2} + 0 = \frac{\sqrt{3} + 1}{2}$
$t a n 45 \degree + c o t 45 \degree = 1 + 1 = 2$

JEE Advanced Numerical:

$s i n (\frac{π}{6}) + c o s (\frac{π}{4}) + t a n (\frac{π}{3}) = \frac{1}{2} + \frac{1}{\sqrt{2}} + \sqrt{3} = \frac{\sqrt{2} + \sqrt{3} + 1}{\sqrt{2}}$
$s i n^{2} (\frac{π}{4}) + c o s^{2} (\frac{π}{3}) + t a n^{2} (\frac{π}{6}) = {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{3}})}^{2} = 1$
$s e c (\frac{π}{3}) - c o s e c (\frac{π}{6}) + c o t (\frac{π}{4}) = 2 - \frac{2}{\sqrt{3}} + 1 = 3 - \frac{2}{\sqrt{3}}$
$t a n (\frac{π}{4}) + c o t (\frac{π}{3}) - s e c (\frac{π}{6}) = \sqrt{3} + \frac{1}{\sqrt{3}} - \sqrt{3} = 0$
$s i n 3 θ + c o s 3 θ = t a n^{2} θ = \frac{\sin^{2} θ}{\cos^{2} θ}$
$s i n (2 θ) + s i n θ = 1 + c o s θ$
$c o s (3 θ) + c o s θ = 1 + s i n 2 θ$
$t a n (3 θ) + t a n θ = s e c^{2} θ$
$s i n^{4} θ + c o s^{4} θ = 1 - 2 s i n^{2} θ c o s^{2} θ$
$s i n^{6} θ + c o s^{6} θ = 1 - 3 s i n^{2} θ c o s^{2} θ$

CBSE Board Exam Numerical:

$\sin 30 \degree = \frac{1}{2}$
$c o s 45 \degree = \frac{1}{\sqrt{2}}$
$t a n 60 \degree = \sqrt{3}$
$s i n^{2} 45 \degree + c o s^{2} 45 \degree = {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$s e c 60 \degree - c o s e c 60 \degree = 2 - \frac{2 \sqrt{3}}{3} = \frac{2 (3 - \sqrt{3})}{3}$
$c o t 45 \degree = \frac{c o s 45 \degree}{s i n 45 \degree} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$
$s i n 30 \degree + c o s 30 \degree = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$
$t a n 60 \degree - c o t 30 \degree = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} (3 - 1)}{\sqrt{3}} = \frac{\sqrt{3} (2)}{\sqrt{3}} = 2$
$s i n 45 \degree + c o s 45 \degree = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$
$s e c 45 \degree - c o s e c 45 \degree = \sqrt{2} - \sqrt{2} = 0$