Class 10 trigonometry Problems with Solutions

$cosec A = \frac {p}{a}$

$cosec^2 A = \frac {p^2}{a^2}$

Similarly

$ cot A = \frac {q}{b}$
$cot^2 A = \frac {q^2}{b^2}$
Now
$cosec^2 A - cot^2 A=1$
$\frac {p^2}{a^2} - \frac {q^2}{b^2} =1$

$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
$x^ -y^2$
$=(a^2 -b^2)(sec^2 \theta -tan^2 \theta)$
$=(a^2 -b^2)$

$cosec A -cot A = q$
$\frac {1}{sin A} - \frac {cos A}{sin A} =q$
$ \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$

$x = p sec \alpha cos \beta$
$\frac {x}{p} = sec \alpha cos \beta$
Squaring
$\frac {x^2}{p^2}= sec^2 \alpha cos^2 \beta$ -(1)

$y = q sec \alpha sin \beta$
$\frac {y}{q} =sec \alpha sin \beta$
Squaring
$\frac {y^2}{q^2} = sec^2 \alpha sin^ \beta$ --(2)
Adding (1) and (2)
$\frac {x^2}{p^2} + \frac {y^2}{q^2} =sec^2 \alpha (cos^2 \beta + sin^2 \beta)$
$\frac {x^2}{p^2} + \frac {y^2}{q^2} =sec^2 \alpha$

$z = r tan \alpha$
$ \frac {z}{r} = tan \alpha$
Squaring
$\frac {z^2}{r^2}= tan^2 \alpha$

Now
$sec^2 \alpha - tan^2 \alpha = 1$
$\frac {x^2}{p^2} + \frac {y^2}{q^2} -\frac {z^2}{r^2} =1$

$tan A + sin A =m$ --(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
$m^ -n^2 = 4 \sqrt {mn}$

LHS
$sin A(1 + tan A) + cos A(1 + cot A)$
$= sin A( 1 + \frac {sin A}{cos A}) + cos A ( 1 + \frac {cos A }{sin A})$
$= sin A \frac {sin A + cos A}{cos A} + cos A \frac {sinA + cos A}{sin A}$
$= (sin A + cos A) ( \frac {sin A }{cos A} + \frac {cos A}{sin A})$
$= (sin A + cos A) (\frac {sin^2 A + cos^2 A}{sin A cos A})$
$=\frac {sin A + cos A}{sin A cos A}$
$=sec A + cosec A$

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
$= \sqrt {(tan A + cot A)^2}$
$=tan A + cot A$

$xsin^3 A + ycos^3 A = sin A cos A$
$(xsin A) sin^2A + (ycosA)cos^2 A = sin A cos A$
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
$x^2 + y^2 = sin^2 A + cos^2 A =1$