Trigonometric Functions questions

$\sqrt 3 cosec 20^0 - sec 20^0$
$=\frac {\sqrt 3}{\sin 20} - \frac {1}{\cos 20}$
$=\frac {\sqrt 3 \cos 20 - \sin 20}{sin 20 cos 20}$
$=4\frac {(\sqrt 3/2) \cos 20 - (1/2) \sin 20}{2sin 20 cos 20}$
$=4 \frac {sin 60 cos 20 - cos60 sin 20}{sin 40}$
$=4 \frac {sin (60 0 20 )}{sin 40}$
= 4

$(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 + \cos \frac {5\pi}{8})(1 + \cos \frac {7\pi}{8})$
$=(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 + \cos (\pi - \frac {3\pi}{8}))(1 + \cos ( \pi - \frac {\pi}{8}))$
$=(1 + \cos \frac {\pi}{8})(1 + \cos \frac {3\pi}{8})(1 - \cos \frac {\pi}{8})(1 - \cos \frac {3\pi}{8})$
$=(1 - \cos^2 \frac {\pi}{8})(1 - \cos^2 \frac {3\pi}{8})$
$=\sin^2 \frac {\pi}{8} \sin^2 \frac {3\pi}{8}$
Now
$\cos 2A = \cos^2 A - \sin^2 A = 1 - 2 \sin^2 A$
Therefore
$=\frac {1}{4} (1 - \cos \frac {\pi}{4})(1 - \cos \frac {3\pi}{4})$
$=\frac {1}{4} (1 - \cos \frac {\pi}{4})(1 - \cos ( \pi - \frac{\pi}{4}))$
$=\frac {1}{4} (1 - \cos \frac {\pi}{4})(1 + \cos \frac {\pi}{4})$
$=\frac {1}{4} (1 - \cos^2 \frac {\pi}{4})$
$=\frac {1}{8}$

$a \cos \theta + b \sin \theta = m$
Squaring both sides
$a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta=m^2$ -(1)
$a \sin \theta - b \cos \theta = n$
Squaring both the sides
$a^2 \sin^2 \theta + b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta=n^2$ --(2)
Adding (1) and (2)
$a^2 (\cos^2 \theta + \sin^2 \theta) + b^2 (\sin^2 \theta + \cos^2 \theta ) = m^2 + n^2$
or
$a^2+ b^2= m^2+ n^2$

$tan (\alpha + \beta) = \frac {tan (\alpha) + tan (\beta)}{1 - tan (\alpha) tan (\beta)}$
Substituting the values, we have
$tan (\alpha + \beta) =1$
$tan (\alpha + \beta) = tan (\frac {\pi}{4})$
$\alpha + \beta = \frac {\pi}{4}$

$5 cos \theta + 3 cos(\theta + \frac {\pi}{3}) + 3$
$=5 cos \theta + 3 cos \theta cos \frac {\pi}{3} - 3 sin \theta sin \frac {\pi}{3} + 3$
$=\frac {1}{2} [ 13 cos \theta - 3 \sqrt 3 sin \theta ] + 3$
Let $13= r cos \alpha$ and $3 \sqrt 3 = r sin \alpha$
$r^2 =13^2 + (3 \sqrt 3)^2 = 196$
$r=14$
So equation becomes
$=\frac {1}{2} [ 14 cos (\theta + \alpha) ] + 3$
$= 7 cos (\theta + \alpha) + 3$
Now
$-1 \leq cos (\theta + \alpha) \leq 1$
$-7 \leq 7cos (\theta + \alpha) \leq 7$
$-4 \leq 7cos (\theta + \alpha) + 3 \leq 14$
Hence Proved

$\frac {\pi}{2} \leq x \leq \pi$
$\frac {\pi}{4} \leq x/2 \leq \frac {\pi}{2}$
So, x/2 lies in First quadrant
Now As x is in second quadrant,cos x will be negative
$cos x= - \frac {1}{\sqrt {1 + tan^2x}}= - \frac {3}{5}$
Now Since x/2 lies in First quadrant, sin (x/2),cos (x/2),tan (x/2) will be positive
$cos \frac {x}{2} = \sqrt { \frac {1 + cos x}{2}}= \frac {1}{\sqrt 5}$
$sin \frac {x}{2} = \sqrt { \frac {1 - cos x}{2}}= \frac {2}{\sqrt 5}$
$tan \frac {x}{2} = \sqrt { \frac {1 - cos x}{1 + cos x}}= 2$

Use the identities
$\cos (2 \theta)=\frac {1- \tan^2 \theta}{1 + \tan^2 \theta}$
and
$\sin (2 \theta)=\frac {2 \tan \theta}{1 + \tan^2 \theta}$
Substituting these values,creating quadratic in tan,then the sum of roots i.e $$\tan (\alpha) + \tan (\beta)$ can be easily derived

$cot 7 \frac {1}{2} ^0$
$= \frac {cos 7 \frac {1}{2}^0}{sin 7 \frac {1}{2} ^0} = \frac {cos 7 \frac {1}{2}^0 cos 7 \frac {1}{2}^0}{sin 7 \frac {1}{2} ^0 cos 7 \frac {1}{2}^0}$
$=\frac {2 cos^2 7 \frac {1}{2}^0}{2 sin 7 \frac {1}{2} ^0 cos 7 \frac {1}{2}^0}=\frac {1 + cos 15^0}{sin 15^0}$
$=\frac {1 + cos (60 -45)}{sin (60-45)}$
Now applying cos(A-B) and sin (A-B) formula and substituting the known values of 60 and 45 angles, we get
$=\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6$

Now
$sin \alpha + sin \beta = -a$ -(1)
$cos \alpha + cos \beta = -c$ -(2)
For (1)
$2 \sin (\frac {\alpha + \beta}{2}) \cos (\frac {\alpha - \beta}{2})=-a$
for (2)
$2 \cos (\frac {\alpha + \beta}{2}) \cos (\frac {\alpha - \beta}{2})=-c$
Dividing both
$\tan (\frac {\alpha + \beta}{2})= \frac {a}{c}$
Now
$sin (\alpha + \beta)= \frac {2 tan (\frac {\alpha + \beta}{2})}{2 + tan^2 (\frac {\alpha + \beta}{2})}= \frac {2ac}{a^2 +c^2}$

$a tan \theta + b sec \theta = c$
$a tan \theta -c = -bsec \theta$
squaring both the sides
$(a tan \theta c)^2= b^2 sec^2 \theta$
$(a tan \theta c)^2= b^2(1 + tan^2 \theta)$
Rearranging
$(a^2 b^2) tan^2 \theta 2ac tan\theta + c^2 b^2=0$
Now $\alpha$ and $\beta$ are solutions, so
$tan \alpha + tan \beta= \frac {2ac}{a^2 -b^2}$
$tan \alpha tan \beta= \frac {c^2 -b^2}{a^2 -b^2}$
Now
$tan (\alpha + \beta) = \frac {tan (\alpha) + tan (\beta)}{1 - tan (\alpha) tan (\beta)}$
Substituting the values, we have
$tan (\alpha + \beta) =\frac {2ac}{a^2 c c^2}$