Question 4.
if $tan (\alpha)= \frac {m}{m+1}$ and $tan (\beta)= \frac {1}{2m+1}$, Prove that $\alpha + \beta = \frac {\pi}{4}$ Solution
$tan (\alpha + \beta) = \frac {tan (\alpha) + tan (\beta)}{1 - tan (\alpha) tan (\beta)}$
Substituting the values, we have
$tan (\alpha + \beta) =1$
$tan (\alpha + \beta) = tan (\frac {\pi}{4})$
$\alpha + \beta = \frac {\pi}{4}$
Question 5.
Prove that $5 cos \theta + 3 cos(\theta + \frac {\pi}{3}) + 3$ lies between -4 and 10 Solution
$5 cos \theta + 3 cos(\theta + \frac {\pi}{3}) + 3$
$=5 cos \theta + 3 cos \theta cos \frac {\pi}{3} - 3 sin \theta sin \frac {\pi}{3} + 3$
$=\frac {1}{2} [ 13 cos \theta - 3 \sqrt 3 sin \theta ] + 3$
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
Question 6.
if $ \frac {\pi}{2} \leq x \leq \pi $ and tan x =-4/3 , Find the value of sin (x/2),cos (x/2),tan (x/2) Solution
$ \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
$cos \frac {x}{2} = \sqrt { \frac {1 + cos x}{2}}= \frac {1}{\sqrt 5}$
$sin \frac {x}{2} = \sqrt { \frac {1 - cos x}{2}}= \frac {2}{\sqrt 5}$
$tan \frac {x}{2} = \sqrt { \frac {1 - cos x}{1 + cos x}}= 2$
Question 7. Prove that
a. $\sin 12^0 \sin 48^0 \sin 54^0= \frac {1}{8}$
b. $\sin 20^0 \sin 40^0 \sin 60^0 \sin 80^0= \frac {3}{16}$
Question 8.
If $a \cos (2 \theta) + b \sin (2\theta) = c$ has $\alpha$ and $\beta$ as its roots, then prove that
$\tan (\alpha) + \tan (\beta) =\frac {2b}{a+c}$ Solution
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
$=\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6$
Question 10.
if $\alpha$ and $\beta$ are the solutions of $sin^2 x + a sin x+ b=0$ as well as that of $cos^2 x + c cos (x) +d=0$,then prove that
$sin (\alpha + \beta) = \frac {2ac}{a^2 + c^2}$ Solution
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
$sin (\alpha + \beta)= \frac {2 tan (\frac {\alpha + \beta}{2})}{2 + tan^2 (\frac {\alpha + \beta}{2})}= \frac {2ac}{a^2 +c^2}$
Question 11.
If $\alpha$ and $\beta$ are the solutions of the equation $a tan \theta + b sec \theta = c$,
then show that $tan (\alpha + \beta) = \frac {2ac}{a^2 - c^2}$ Solution
$a tan \theta + b sec \theta = c$
$a tan \theta -c = -bsec \theta$
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
$tan (\alpha + \beta) =\frac {2ac}{a^2 c c^2}$
Question 12.
Show that
$\frac {sec^2 \theta - tan \theta}{sec^2 \theta + tan \theta}$ lies between 1/3 and 3
Question 13.
In any $\Delta ABC$ , if $a^2, b^2 ,c^2$ are in A.P,Prove that cot A,cot B and Cot C are in A.P
