For positive sign
Now $Sec X + Tan X = (a + \frac {1}{4a}) + (a - \frac {1}{4a})$
= 2a
For negative sign
Now $Sec X + Tan X = (a + \frac {1}{4a}) - (a - \frac {1}{4a})$
$ = \frac {1}{2a}$
Question 2.
If $Sin A+ Sin^2 A= 1$, then find the value of $(cos^2 A+cos^4 A)$. Solution
$Sin A+ Sin^2 A= 1$
$sin A = 1 -sin^2 A$
$sin A =cos^2 A$
Now
$(cos^2 A+cos^4 A)$
$=(sinA + sin^2 A)$
$=1$
Question 3
If $x(cos A) - y(sin A) = a$, $x(sin A) + y(cos A) = b$, the tick mark whichever option is correct
a. $x^2 -y^2 = a^2 -b^2$
b. $x^2 +y^2 = a^2 +b^2$
c. $x^2 +y^2 = a^2 - b^2$
d. $x^2 -y^2 = a^2 + b^2$ Solution
$x(cos A) - y(sin A) = a$, $x(sin A) + y(cos A) = b$
Squaring both the equation and adding
$x^2(cos^2 A + sin^2 A) + y^2(cos^2 A + sin^2 A) -2xy sin A cos A + 2xy sin A cos A=a^2 + b^2$
Now as $ sin^2 A + cos^2 A =1$
$x^2 + y^2 =a^2 + b^2$
Answer (b)
Question 4
If $tan 2A = cot (A - 18^{\circ})$, where 2A is an acute angle. Find the value of A. Solution
As Cot ( 90 - x ) = tan x
According to the problem given
$tan 2A = cot (A - 18^{\circ})$
$
Cot ( 90 - 2A ) = cot ( A - 18^{\circ})$
or
90 - 2A = A - 18
3A =108
A=36 °
Question 5
If $tan (A+B) = \sqrt {3}$ and $tan (A - B) = \frac {1}{\sqrt {3}}$ Find the value of A and B. Solution
Question 11
If A and B acute angles such that $tan A = \frac {1}{2}$ , $tan B = \frac {1}{3}$ and
$tan (A + B) =\frac { tan A + tan B}{1- tan A tan B}$, find A + B. Solution
$tan (A + B) =\frac { tan A + tan B}{1- tan A tan B}$
$=\frac { 1/2 + 1/3}{1 - 1/2 \times 1/3} = 1$
$tan (A + B) =tan 45$
A + B =45
Question 12
Prove that
a. $tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ} = 1$
b. $tan 1^{\circ} tan 2^{\circ} tan 3^{\circ} .... tan 89^{\circ} = 1$
c. $cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ} = 0$ Solution
a. $tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ} = 1$
LHS
$tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ}$
$=tan 10^{\circ} tan 15^{\circ} tan (90^{\circ} -15 ^{\circ}) tan (90^{\circ} -10^{\circ})$
$=tan 10^{\circ} tan 15^{\circ} cot 15^{\circ} cot 10^{\circ}$
$=tan 10^{\circ} tan 15^{\circ} \frac {1}{tan 15^{\circ}} \frac {1}{tan 10^{\circ}}$
$=1$
c. $cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ}$
LHS
Since cos 90=1
$cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ}=1$
Question 13
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
a. $ \frac {cos A}{1 + sin A} + \frac {1 + sin A}{cos A} = 2 sec A$
b. $ \frac {1 + sec A}{sec A} = \frac {sin^2 A}{1- cos A}$
c. $ \sqrt { \frac {1+ sin A}{1- sin A} } = sec A + tan A$ Solution
a. LHS
$ \frac {cos A}{1 + sin A} + \frac {1 + sin A}{cos A}$
$= \frac {cos^2 A + 1 +sin^2 A + 2 sinA}{cosA(1+ SinA)}$
$=\frac {2+ 2 sin A}{cos A(1+sin A)}$
$= \frac {2}{cos A}$
$=2 sec A$
=RHS
RHS
$\frac {sin^2 A}{1- cos A}$
$= \frac {1 - cos^2 A}{1- cos A}$
$=1 + cos A$
c. LHS
$ \sqrt { \frac {1+ sin A}{1- sin A} }$
$=\sqrt { \frac {1+ sin A}{1- sin A} \times \frac {1+ sin A}{1 +sin A} }$
$=\sqrt { \frac {(1+sin A)^2}{cos^2 A}}$
$= \frac {1+ sin A}{cos A}$
$=sec A + tan A$
Question 14
In a $\Delta ABC$ right angled at C, if $tan A = \frac {1}{\sqrt {3}}$ find the value of
$sin A cos B + cos A sin B$. Solution
$tan A = \frac {1}{\sqrt {3}}$
$tan A = tan 30$
A=30°
Now C=90°
B=60°
Then
$sin A cos B + cos A sin B$
$=sin 30 cos 60 + cos 30 sin 60$
$ =\frac {1}{4} + \frac {3}{4}$
$=1$
Question 15
If $ sec \theta - tan \theta = x$, show that:
$sec \theta = \frac {1}{2} [x + \frac {1}{x}]$
$tan \theta =\frac {1}{2} [\frac {1}{x} -x]$ Solution
$ sec \theta - tan \theta = x$ - (1)
$(sec \theta - tan \theta) \times \frac {sec \theta - tan \theta}{sec \theta + tan \theta} = x$
$\frac {sec^2 \theta - tan^2 \theta}{sec \theta + tan \theta}=x$
$\frac {1}{sec \theta + tan \theta}=x$
$sec \theta + tan \theta = \frac {1}{x}$ -(2)
Adding (1) and (2)
$2 sec \theta = x + \frac {1}{x}$
$sec sec \theta=\frac {1}{2} [x + \frac {1}{x}]$
subtracting (1) from (2)
$2 tan \theta= \frac {1}{x} -x$
$tan \theta =\frac {1}{2} [\frac {1}{x} -x]$
Question 16.
If $tan \theta = \frac {12}{5}$
Find the value $ \frac {1+ sin \theta }{1 -sin \theta }$ Solution
$tan \theta = \frac {12}{5}$
$\frac {P}{B} =\frac {12}{5}$
So P=12k and B=5k
$H^2 =P^2 + B^2= (13k)^2$
H=13K
$ sin \theta = \frac {12}{13}$
Now
$ \frac {1+ sin \theta }{1 -sin \theta }$
$= \frac {25}{1} = 25$
Question 17
If $sin \theta + cos \theta = \sqrt {2} cos (90 - \theta)$,find $cot \theta$ Solution
$sin \theta + cos \theta = \sqrt {2} cos (90 - \theta)$
$sin \theta + cos \theta = \sqrt {2} sin \theta$
$cos \theta= sin \theta ( \sqrt {2} -1)$
$cot \theta=\sqrt {2} -1$
Question 18 Prove that
If $ tan^2 \theta = 1 -p^2 $, then prove that $sec \theta + tan^3 \theta cosec \theta = (2 - p^2) ^ {3/2}$. Solution
