Question 1.
Given
$a cosec A = p$ and $b cot A = q$,
then prove that
$\frac {p^2}{a^2} - \frac {q^2}{b^2} =1$ Solution
$cosec A = \frac {p}{a}$
$cosec^2 A = \frac {p^2}{a^2}$
Similarly
$ cot A = \frac {q}{b}$
$cot^2 A = \frac {q^2}{b^2}$
Now
$cosec^2 A - cot^2 A=1$
$\frac {p^2}{a^2} - \frac {q^2}{b^2} =1$
Question 2.
Prove the following trigonometric identities:
a.$cos^2 \theta (1 + tan^2 \theta) = 1$
b. $cos^2 \theta + \frac {1}{1+ cot^2 \theta} =1$
c. $ \frac {1}{1+ sin \theta } + \frac {1}{1-sin \theta } = 2 sec^2 \theta$
d. $cosec^2 \theta + sec^2 \theta =cosec^2 \theta sec^2 \theta$
e.$tan \theta - cot \theta = \frac {2sin^2 \theta -1}{sin \theta cos \theta}$
Question 3.
If A, B, C are interior angles of $\Delta ABC$, show that
$cosec ^2 (\frac {B+C}{2} ) - tan ^2 \frac {A}{2} =1$
Question 4.
If $x = a sec \theta + b tan \theta $ and $y=a tan \theta + b sec \theta $ ,prove that
$x^2 -y^2 = a^2 -b^2$ Solution
$x = a sec \theta + b tan \theta $
$x^2 =a^2 sec^2 \theta + b^2 tan^2 \theta + ab sec \theta tan \theta$
Similarly
$y=a tan \theta + b sec \theta $
$y^2 = a^2 tan^2 \theta + b^2 sec^2 \theta + ab sec \theta tan \theta$
Now
$x^ -y^2$
$=(a^2 -b^2)(sec^2 \theta -tan^2 \theta)$
$=(a^2 -b^2)$
Question 5
Prove the following identities:
i. $2(sin^6 \theta + cos^6 \theta ) -3 (sin^4 \theta + cos^4 \theta ) + 1=0$
ii. $sin^6 \theta + cos^6 \theta + 2 sin^2 \theta cos ^2 theta =1$
iii. $(sin^8 \theta -cos^8 \theta = (sin^2 \theta -cos^2 \theta) (1 -2sin^2 \theta cos^2 \theta )$
Question 6.
If $cosec A -cot A = q$, then show that
$ \frac {q^2 -1}{q^2 + 1} + cos A =0$ Solution
$cosec A -cot A = q$
$\frac {1}{sin A} - \frac {cos A}{sin A} =q$
$ \frac {1-cos A }{sin A}=q$
$ \frac {(1-cos A)^2}{sin^2 A} = q^2$
$\frac {1- cos A}{1+ cos A} = q^2$
$ \frac {q^2 -1}{q^2 + 1} + cos A =0$
Question 7.
If $x = p sec \alpha cos \beta$, $y = q sec \alpha sin \beta$ and $z = r tan \alpha$, then show that
$\frac {x^2}{p^2} + \frac {y^2}{q^2} -\frac {z^2}{r^2} =1$ Solution
$x = p sec \alpha cos \beta$
$\frac {x}{p} = sec \alpha cos \beta$
Squaring
$\frac {x^2}{p^2}= sec^2 \alpha cos^2 \beta$ -(1)
Question 8. Prove that
$ (1+ cot \theta) (1 + tan \theta + sec \theta )=2$
Question 9. Prove that
cot4 A – 1 = cosec4 A – 2cosec2 A Question 10.
$ \frac {sin A -cos A +1}{sin A + cos A -1} = \frac {1}{sec A -tan A}$ Question 11.
if $tan A + sin A =m$ and $tan A -sin A =n$
Prove that Solution
$m^ -n^2 = 4 \sqrt {mn}$
$tan A + sin A =m$ --(1)
$tan A -sin A =n$ --(2)
Squaring and subtracting
$m^2 -n^2 = tan^2 A + sin^2 A +2 tan A sin A - tan^2 A -sin^2 A + 2tan A sin A$
$m^2 -n^2 = 4tan A sin A$
Multiplying 1 and 2
$tan^2 A -sin^2 A = mn$
$sin^2 A ( \frac {1}{cos^2 A} -1) = mn$
$sin^2 A \frac {1-cos^2 A}{cos^2 A}=mn$
$ tan^2 A (1-cos^2A) =mn$
$tan^2 A sin^2 A =mn$
$tan A sin A = \sqrt {mn}$
Hence
$m^ -n^2 = 4 \sqrt {mn}$
Question 12
$ sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A$ Solution
LHS
$sin A(1 + tan A) + cos A(1 + cot A)$
$= sin A( 1 + \frac {sin A}{cos A}) + cos A ( 1 + \frac {cos A }{sin A})$
$= sin A \frac {sin A + cos A}{cos A} + cos A \frac {sinA + cos A}{sin A}$
$= (sin A + cos A) ( \frac {sin A }{cos A} + \frac {cos A}{sin A})$
$= (sin A + cos A) (\frac {sin^2 A + cos^2 A}{sin A cos A})$
$=\frac {sin A + cos A}{sin A cos A}$
$=sec A + cosec A$
Question 13
$ \frac {tan A}{1- cot A} + \frac {cot A}{1-tan A} = 1 + tan A + cot A$ Question 14
$\sqrt {sec^2 A + cosec^2 A} = (tan A + cot A)$ Solution
LHS
$\sqrt {sec^2 A + cosec^2 A}$
$= \sqrt {1 + tan^2 A + 1 + cot^2 A}$
$= \sqrt { tan^2 A + cot^2 A + 2}$
$= \sqrt { tan^2 A + cot^2 A + 2 tan A cot A}$ as tanA cot A=1
$= \sqrt {(tan A + cot A)^2}$
$=tan A + cot A$
Question 15
if $xsin^3 A + ycos^3 A = sin A cos A$
and $x sin A = y cos A$
Prove that
$x^2 + y^2 =1$ Solution
$xsin^3 A + ycos^3 A = sin A cos A$
$(xsin A) sin^2A + (ycosA)cos^2 A = sin A cos A$
Now $x sin A = y cos A$
$(xsin A) sin^2A + (x sin A)cos^2 A = sin A cos A$
$xsinA (sin^2 A + cos^2 A) =sin A cos A$
$x=cos A$
Now
$x sin A = y cos A$
$cos A sin A = y cos A$
$y=sin A$
Now
$x^2 + y^2 = sin^2 A + cos^2 A =1$
