Trigonometry Important Questions for Class 10 Maths

$x(cos A) - y(sin A) = a$, $x(sin A) + y(cos A) = b$
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)

$cos A=\frac {1}{2}$
$cos A = cos 60$
A=60°

$sin B =\frac {1}{2}$
$sin B = sin 30$
B=30°
Therefore,
A+ B= 90°

Answer (c)

$sin (X + Y) =1$
$sin (X+ Y) = sin90$
X+ Y=90

$cos (X - Y) =1$
$cos( X-Y) =cos 0$
X-Y =0
Solving these we get
X = Y = 45°

Answer (c)

$sec X= a + \frac {1}{4a}$
$Tan X = \sqrt {Sec^2 X - 1}$
$= \sqrt {(a + \frac {1}{4a})^2 - 1}$
$= \sqrt {(a^2 + \frac {1}{16}a^2 + \frac {1}{2}) - 1}$
$= \sqrt {a^2 + \frac {1}{16}a^2 - \frac {1}{2}}$
$= \sqrt {(a - \frac {1}{4a})^2}$
$= \pm (a - \frac {1}{4a})$

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}$

$Sin A+ Sin^2 A= 1$
$sin A = 1 -sin^2 A$
$sin A =cos^2 A$

Now
$(cos^2 A+cos^4 A)$
$=(sinA + sin^2 A)$
$=1$

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 °

$tan (A+B) = \sqrt {3}$
$tan (A+B) = tan 60$
A+B=60 --(1)

$tan (A - B) = \frac {1}{\sqrt {3}}$
$tan (A - B) = tan 30$
A-B=30 --(2)
Solving 1 and 2
A=45 and B=15

$sin (A+B) = 1$
$sin (A+B) = sin 90^{\circ}$
A+B =90 -(1)

Similarly
$cos (A-B) = \frac {\sqrt {3}}{2}$
$cos (A-B) =cos 30 ^{\circ}$
A-B =30 -(2)
Solving (1) and (2)
$A=60^{\circ})$ and $B=30^{\circ}$

$sin \theta - cos \theta = 0$
$sin \theta = cos \theta $
$\frac {sin \theta}{cos \theta}=1$
$tan \theta=1$
$\theta =45^{\circ}$
Now
$(sin^4 \theta + cos^4 \theta)$
$=\frac {1}{4} + \frac {1}{4} $
$=\frac {1}{2}$

$sec \theta + tan \theta =p$
$ \frac {1}{cos\theta} + \frac {sin \theta}{cos \theta} = p$
$ \frac {sin \theta +1}{cos \theta} = p$

Squaring both the sides
$ \frac { (sin \theta + 1)^2}{cos^2 \theta}= p^2$
$ \frac { (sin \theta + 1)^2}{1 -sin^2 \theta}= p^2$
$ \frac { 1 + sin \theta}{ 1 - sin \theta} = p^2$

or $sin \theta = \frac {p^2 -1}{p^2 + 1} $
Now
$ cosec \theta= \frac {1}{sin \theta} = \frac {p^2 +1}{p^2 - 1}$

$tan (A + B) =\frac { tan A + tan B}{1- tan A tan B}$
$=\frac { 1/2 + 1/3}{1 - 1/2 \times 1/3} = 1$
$tan (A + B) =tan 45$
A + B =45

a. $tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ} = 1$
LHS
$tan 10^{\circ} tan 15^{\circ} tan 75^{\circ} tan 80^{\circ}$
$=tan 10^{\circ} tan 15^{\circ} tan (90^{\circ} -15 ^{\circ}) tan (90^{\circ} -10^{\circ})$
$=tan 10^{\circ} tan 15^{\circ} cot 15^{\circ} cot 10^{\circ}$
$=tan 10^{\circ} tan 15^{\circ} \frac {1}{tan 15^{\circ}} \frac {1}{tan 10^{\circ}}$
$=1$

b. $Tan 1^{\circ}.tan 2^{\circ}.tan 3^{\circ} ......tan 89 ^{\circ}$
$=tan 1^{\circ}.tan 2^{\circ}.tan 3^{\circ}.....tan (90^{\circ}-2 ^{\circ}).tan (90^{\circ}-1 ^{\circ}) $
$=tan 1^{\circ}.tan 2^{\circ}.tan 3^{\circ}.....cot2 ^{\circ} cot 1 ^{\circ}$
=1

c. $cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ}$
LHS
Since cos 90=1
$cos 1^{\circ} cos 2^{\circ} cos3^{\circ} .... cos 180^{\circ}=1$

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

b. LHS
$ \frac {1 + sec A}{sec A}$
$=\frac {1 + \frac {1}{cos A}}{\frac {1}{cos A}}$
$= 1 + cosA$

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} }$
$=\sqrt { \frac {1+ sin A}{1- sin A} \times \frac {1+ sin A}{1 +sin A} }$
$=\sqrt { \frac {(1+sin A)^2}{cos^2 A}}$
$= \frac {1+ sin A}{cos A}$
$=sec A + tan A$

$tan A = \frac {1}{\sqrt {3}}$
$tan A = tan 30$
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}$
$=1$

$ 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)
$2 tan \theta= \frac {1}{x} -x$
$tan \theta =\frac {1}{2} [\frac {1}{x} -x]$

$tan \theta = \frac {12}{5}$
$\frac {P}{B} =\frac {12}{5}$
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 }$
$= \frac {25}{1} = 25$

$sin \theta + cos \theta = \sqrt {2} cos (90 - \theta)$
$sin \theta + cos \theta = \sqrt {2} sin \theta$
$cos \theta= sin \theta ( \sqrt {2} -1)$
$cot \theta=\sqrt {2} -1$

$ tan^2 \theta = 1 -p^2 $
LHS $=sec \theta + tan^3 \theta cosec \theta $
Now as $sec^2 \theta =tan^2 \theta + 1$ and $cosec^2 \theta -cot^2 \theta =1$
$= \sqrt {1+tan^2 \theta}+tan^2 \theta \times tan \theta \times \sqrt {1+cot^2 \theta}$
$=\sqrt {1+(1-p^2)} +(1-p^2) \times \sqrt {1-p^2 } \times \sqrt {1 + \frac {1}{1-p^2}}$
$= \sqrt {2-p^2}+ (1-p^2) \times \sqrt {1-p^2} \times \sqrt {\frac {2-p^2}{1-p^2}}$
$=\sqrt {2-p^2} +(1-p^2) \sqrt {(2-p^2}$
$=\sqrt {2-p^2} \times (1+1-p^2)$
$=\sqrt {2-p^2} \times (2-p^2)$
$=(2-p^2)^{3/2}$

(i) 3/5
(ii)2
(ii) 5/12
(iv) 2

(i)False
(ii)True
(ii)False
(iv) false