Question 13 Marks
If $A$ and $B$ are complementary angles, prove that:
$cosec^2A + cosec^2B = cosec^2A ~cosec^2B$
AnswerSince, A and B are complementary angles, A + B = 90°
$cosec^2A + cosec^2B$
$= cosec^2A + (cosec(90^\circ - A))^2$
$= cosec^2A + sec^2A$
$=\frac{1}{\sin ^2 A}+\frac{1}{\cos ^2 A}$
$=\frac{\cos ^2 A+\sin ^2 A}{\sin ^2 A \cos ^2 A}$
$=\frac{1}{\sin ^2 A \cos ^2 A}$
$= cosec^2A (sec(90A^\circ - B))^2$
$= cosec^2A ~cosec^2B$
View full question & answer→Question 23 Marks
Prove that:
$\frac{1}{1+\sin \left(90^{\circ}-A\right)}+\frac{1}{1-\sin \left(90^{\circ}-A\right)}=2 \sec ^2\left(90^{\circ}-A\right)$
Answer$\frac{1}{1+\sin \left(90^{\circ}-A\right)}+\frac{1}{1-\sin \left(90^{\circ}-A\right)}$
$=\frac{1}{1+\cos A}+\frac{1}{1-\cos A}$
$=\frac{1-\cos A+1+\cos A}{(1+\cos A)(1-\cos A)}$
$=\frac{2}{1-\cos ^2 A}$
$=2 \operatorname{cosec}^2 A $
$ =2 \sec ^2\left(90^{\circ}-A\right)$
View full question & answer→Question 33 Marks
Prove that:
$\frac{1}{1+\cos \left(90^{\circ}-A\right)}+\frac{1}{1-\cos \left(90^{\circ}-A\right)}=2 \operatorname{cosec} 2\left(90^{\circ}-A\right)$
Answer$\frac{1}{1+\cos \left(90^{\circ}-A\right)}+\frac{1}{1-\cos \left(90^{\circ}-A\right)}$
$=\frac{1}{1+\sin A}+\frac{1}{1-\sin A}$
$\frac{1-\sin A+1+\sin A}{(1+\sin A)(1-\sin A)}$
$=\frac{2}{1-\sin ^2 A}$
$=\frac{2}{\cos ^2 A}$
$= 2\sec^2 A$
$= 2 \cos ec^2(90^\circ - A)$
View full question & answer→Question 43 Marks
If $2$ sin $A – 1 = 0$, show that: $Sin\ 3A = 3 \sin A – 4 \sin^3A$
Answer$2 sinA − 1 = 0$
$\Rightarrow sinA = 1/2$
We know $sin30^\circ = 1/2$
$So, A = 30^\circ $
$LHS = \sin 3A = sin90^\circ = 1$
$RHS = 3sinA - 4sin^3A$
$=3 sin30^\circ - 4sin^330^\circ $
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}=1$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 53 Marks
If $\sec A+\tan A=p$, show that:
$\sin A =\frac{p^2-1}{p^2+1}$
Answer$\frac{p^2-1}{p^2+1}$.
$=\frac{(\sec A +\tan A )^2-1}{(\sec A +\tan A )^2+1} $
$ =\frac{\sec ^2 A +\tan ^2 A +2 \tan A \sec A -1}{\sec ^2 A +\tan ^2 A +2 \tan A \sec A +1} $
$=\frac{\tan ^2 A +\tan ^2 A +2 \tan A \sec A }{\sec ^2 A +\sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan ^2 A +2 \tan A \sec A }{2 \sec ^2 A +2 \tan A \sec A }$
$ =\frac{2 \tan A (\tan A +\sec A )}{2 \sec A (\tan A +\sec A )} $
$=\sin A $
View full question & answer→Question 63 Marks
If $x=a \cos \theta$ and $y=b \cot \theta$, show that:
$\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$
Answer$\frac{a^2}{x^2}-\frac{b^2}{y^2} $
$=\frac{a^2}{a^2 \cos ^2 \theta}-\frac{b^2}{b^2 \cot ^2 \theta} $
$=\frac{1}{\cos ^2 \theta}-\frac{\sin ^2 \theta}{\cos ^2 \theta}$
$=\frac{1-\sin ^2 \theta}{\cos ^2 \theta} $
$=\frac{\cos ^2 \theta}{\cos ^2 \theta} $
$=1$
View full question & answer→Question 73 Marks
Evaluate without using trigonometric tables,
$\sin ^2 28^{\circ}+\sin ^2 62^{\circ}+\tan ^2 38^{\circ}-\cot ^2 52^{\circ}+\frac{1}{4} \sec ^2 30^{\circ}$
Answer$\sin ^2 28^{\circ}+\sin ^2 62^{\circ}+\tan ^2 38^{\circ}-\cot ^2 52^{\circ}+\frac{1}{4} \sec ^2 30^{\circ}$
$ =\sin ^2 28^{\circ}+\sin ^2\left(90^{\circ}-28^{\circ}\right)+\tan ^2 38^{\circ}-\cot ^2\left(90^{\circ}-38^{\circ}\right)+\frac{1}{4} \sec ^2 30^{\circ}$
$ =\left(\sin ^2 28^{\circ}+\cos ^2 28^{\circ}\right)+\tan ^2 38^{\circ}-\tan ^2 38^{\circ}+\frac{1}{4} \times\left(\frac{2}{\sqrt{3}}\right)^2 $
$ =1+0+\frac{1}{4} \times \frac{4}{3}$
$ =1+\frac{1}{3}$
$=\frac{4}{3}$
View full question & answer→Question 83 Marks
Prove the identity $(\sin \theta + \cos \theta) (\tan \theta + \cot \theta ) = \sec \theta + \cos ec θ.$
Answer$\text { L.H.S. }=(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)$
$=(\sin \theta+\cos \theta) \times \frac{1}{\sin \theta \cos \theta}\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}$
$=\frac{\sin \theta}{\cos \theta \sin \theta}+\frac{\cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$
$=\sec \theta+\cos e c \theta$
$= R.H.S$
Hence proved.
View full question & answer→Question 93 Marks
Prove that:
$(cosecA - sinA) (secA - cosA) \sec^2A = tanA$
AnswerLHS,
$(co\sec A - \sin A) (\sec A - \cos A) \sec^2A$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \sec ^2 A$
$=\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right) \sec ^2 A$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right) \sec ^2 A$
$=\frac{\sin A}{\cos A}=\tan A= RHS $
View full question & answer→Question 103 Marks
If $0^\circ < A < 90^\circ;$ Find $A,$ if $\frac{\sin A}{\sec A-1}+\frac{\sin A}{\sec A+1}=2$
Answer$\frac{\sin A}{\sec A-1}+\frac{\sin A}{\sec A+1}=2$
$\Rightarrow \frac{\sin A \sec A+\sin A+\sec A \sin A-\sin A}{(\sec A-1)(\sec A+1)}=2$
$\Rightarrow \frac{2 \sin A \sec A}{\sec ^2 A-1}=2$
$\Rightarrow \frac{\sin A \sec A}{\tan ^2 A}=1$
$\Rightarrow \frac{\cos A}{\sin A}=1$
$\Rightarrow \cot A=1$
We know $\cot 45^\circ = 1$
Hence, $A = 45^\circ$
View full question & answer→Question 113 Marks
If $0^\circ < A < 90^\circ;$ Find A, if :
$\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=4$
Answer$\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=4 $
$ \Rightarrow \frac{\cos A+\cos A \sin A+\cos A-\sin A \cos A}{(1-\sin A)(1+\sin A)}=4 $
$ \Rightarrow \frac{2 \cos A}{1-\sin A}=4 $
$ \Rightarrow \frac{2 \cos A}{\cos ^2 A}=4 $
$ \Rightarrow \frac{1}{\cos A}=2 $
$ \Rightarrow \cos A=\frac{1}{2}$
We know $\cos 60=\frac{1}{2}$
Hence, $A=60$
View full question & answer→Question 123 Marks
Find A, if $0^\circ \leq A \leq 90^\circ$ and $2cos^2A+ cosA - 1 = 0$
Answer$2cos^2A+ cosA - 1 = 0$
$\Rightarrow 2cos^2A + 2cosA - cosA - 1 = 0$
$\Rightarrow 2cosA (cosA + 1) - 1(cosA + 1) = 0$
$\Rightarrow (2cosA - 1)(cosa + 1) = 0$
$\Rightarrow cosA = 1/2$ or $cosA = -1$
We know $\cos 60^\circ = 1/2$
We also know that for no value of $A(0^\circ \leq A\leq 90^\circ ), cosA = -1$.
Hence, $A = 60^\circ$
View full question & answer→Question 133 Marks
Prove the following identitie:
$\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}$
Answer$\sqrt{\frac{1-\cos A}{1+\cos A}} $
$ =\sqrt{\frac{1-\cos A}{1+\cos A} \times \frac{1+\cos A}{1+\cos A}} $
$ =\sqrt{\frac{1-\cos ^2 A}{(1+\cos A)^2}} $
$ =\sqrt{\frac{\sin ^2 A}{(1+\cos A)^2}} $
$ =\frac{\sin A}{1+\cos A}$
View full question & answer→Question 143 Marks
Prove the following identitie:
(1 + tanA + secA)(1 + cotA - cosecA) = 2
Answer(1 + tanA + secA) (1 + cotA - cosecA)
=1 + cotA - cosecA + tanA + 1 - secA + secA + cosecA - cosecA secA
$=2+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}-\frac{1}{\sin A \cos A}$
$=2+\frac{\cos ^2 A+\sin ^2 A}{\sin A \cos A}-\frac{1}{\sin A \cos A}$
$=2+\frac{1}{\sin A \cos A}-\frac{1}{\sin A \cos A}$
$=2$
View full question & answer→Question 153 Marks
Prove the following identitie:
$cosec^4A (1 - \cos^4A) - 2 \cot^2A = 1$
Answer$cosec^4A (1 - \cos^4A) - 2 \cot^2A$
$=cosec^4A (1 - \cos^2A) (1 + \cos^2A) - 2 \cot^2A$
$= cosec^4A (\sin^2A) (1 + \cos^2A) - 2 \cot^2A$
$= cosec^2A (1 + \cos^2A) - 2 \cot^2A$
$=\cos e c^2 A+\frac{\cos ^2 A}{\sin ^2 A}-2 \cot ^2 A$
$= cosec^2A + \cot^2A - 2cot^2A$
$= cosec^2A - \cot^2A$
$= 1$
View full question & answer→Question 163 Marks
Prove the following identitie:
$sec^4A (1 - \sin^4A) - 2 \tan^2A = 1$
Answer$sec^4A (1 - \sin^4A) - 2 \tan^2A$
$= sec^4A (1 - \sin^2A) (1 + \sin^2A) - 2tan^2A$
$= sec^4A (\cos^2A) (1 + \sin^2A) - 2tan^2A$
$=\sec ^2 A+\frac{\sin ^2 A}{\cos ^2 A}-2 \tan ^2 A$
$= sec^2A + \tan^2A - 2 \tan^2A$
$= sec^2A - \tan^2A$
$= 1$
View full question & answer→Question 173 Marks
Prove the following identitie:
$\frac{\left(1-2 \sin ^2 A\right)^2}{\cos ^4 A-\sin ^4 A}=2 \cos ^2 A-1$
Answer$\frac{\left(1-2 \sin ^2 A\right)^2}{\cos ^4 A-\sin ^4 A} $
$=\frac{\left(1-2 \sin ^2 A\right)^2}{\left(\cos ^2 A-\sin ^2 A\right)\left(\cos ^2 A+\sin ^2 A\right)}$
$ =\frac{\left(1-2 \sin ^2 A\right)^2}{1-\sin ^2 A-\sin ^2 A} $
$ =\frac{\left(1-2 \sin ^2 A\right)^2}{1-2 \sin ^2 A} $
$ =1-2 \sin ^2 A $
$ =1-2\left(1-\cos ^2 A\right)$
$=2 \cos ^2 A-1$
View full question & answer→Question 183 Marks
If $4 \cos^2 A – 3 = 0$ and $\leq A \leq 90^\circ$, then prove that :
$\sin 3 A = 3 \sin A – 4 \sin^3 A$
Answer$4 \cos^2A − 3 = 0$
$\cos A=\frac{\sqrt{ } 3}{2}$
We know $\cos 30=\frac{\sqrt{3}}{2}$
So, A = 30
(i) LHS = sin 3A = sin 90 = 1
$RHS = 3sin A - 4 \sin^3 A$
$= 3sin\ 30 - 4 \sin^3 30$
$=3 \times \frac{1}{2}-4 \times\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}$
$=1$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 193 Marks
Prove the following identitie:
$\sqrt{\frac{1+\sin A}{1-\sin A}}=\frac{\cos A}{1-\sin A}$
Answer$\sqrt{\frac{1+\sin A}{1-\sin A}}$
$=\sqrt{\frac{1+\sin A}{1-\sin A} \times \frac{1-\sin A}{1-\sin A}} $
$=\sqrt{\frac{1-\sin ^2 A}{(1-\sin A)^2}}$
$ =\sqrt{\frac{\cos ^2 A}{(1+\sin A)^2}} $
$=\frac{\cos A}{1-\sin A}$
View full question & answer→Question 203 Marks
Prove the following identitie:
$\frac{\sin A}{1-\cos A}-\cot A=\operatorname{cosec} A$
Answer$\frac{\sin A}{1-\cos A}-\cot A $
$ =\frac{\sin A}{1-\cos A}-\frac{\cos A}{\sin A}$
$=\frac{\sin ^2 A-\cos A+\cos ^2 A}{(1-\cos A) \sin A} $
$ =\frac{1}{\sin A} $
$ =\operatorname{cosec} A$
View full question & answer→Question 213 Marks
Prove the following identitie:
$\operatorname{cosec} A-\cot A=\frac{\sin A}{1+\cos A}$
Answer$\operatorname{cosec} A-\cot A$
$=\frac{1}{\sin A}-\frac{\cos A}{\sin A} $
$ =\frac{1-\cos A}{\sin A} $
$ =\frac{1-\cos A}{\sin A} \times \frac{1+\cos A}{1+\cos A} $
$ =\frac{1-\cos ^2 A}{\sin A(1+\cos A)}$
$\frac{\sin 2}{\sin A(1+\cos A)}$
$=\frac{\sin A}{1+\cos A}$
View full question & answer→Question 223 Marks
Prove that
$\frac{\cot A-1}{2-\sec ^2 A}=\frac{\cot A}{1+\tan A}$
Answer$\text { L.H.S. }=\frac{\cot A -1}{2-\sec ^2 A } $
$ =\frac{\frac{1}{\tan A }-1}{2-\left(1+\tan ^2 A \right)}$
$=\frac{1-\tan A }{\tan A \left(1-\tan ^2 A \right)} $
$ =\frac{1-\tan A }{\tan A (1-\tan A )(1+\tan A )}$
$ =\frac{1}{\tan A (1+\tan A )}$
$=\frac{1}{\tan A } \times \frac{1}{1+\tan A } $
$ =\frac{\cot A }{1+\tan A }$
$R.H.S.$
Hence proved.
View full question & answer→Question 233 Marks
Prove the following identitie:
$\frac{1}{\cos A+\sin A}+\frac{1}{\cos A-\sin A}=\frac{2 \cos A}{2 \cos ^2 A-1}$
Answer$\frac{1}{\cos A+\sin A}+\frac{1}{\cos A-\sin A}$
$=\frac{\cos A+\sin A+\cos A-\sin A}{(\cos A+\sin A)(\cos A-\sin A)} $
$ =\frac{2 \cos A}{\cos ^2 A-\sin ^2 A} $
$ =\frac{2 \cos A}{\cos ^2 A-\left(1-\cos ^2 A\right)} $
$ =\frac{2 \cos A}{2 \cos ^2 A-1}$
View full question & answer→Question 243 Marks
Prove that
(tanA + cotA) (cosecA - sinA) (secA - cosA) = 1
Answer$(\tan A+\cot A)(\operatorname{cosec} A-\sin A)(\sec A-\cos A)$
$=\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$
$=\left(\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A}\right)\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)$
$=\left(\frac{1}{\sin A \cos A}\right)\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)$
$=1$
View full question & answer→Question 253 Marks
Prove that
$\frac{1}{\sin A-\cos A}-\frac{1}{\sin A+\cos A}=\frac{2 \cos A}{2 \sin ^2 A-1}$
Answer$\frac{1}{\sin A-\cos A}-\frac{1}{\sin A+\cos A} $
$ =\frac{\sin A+\cos A-\sin A+\cos A}{(\sin A-\cos A)(\sin A+\cos A)} $
$ =\frac{2 \cos A}{\sin ^2 A-\cos ^2 A} $
$ =\frac{2 \cos A}{\sin ^2 A-\left(1-\sin ^2 A\right)}$
$ =\frac{2 \cos A}{2 \sin ^2 A-1}$
View full question & answer→Question 263 Marks
Find the value of x, if $cos(2x - 6) = \cos^230^\circ - \cos^260^\circ$
Answer$\cos(2x - 6) = \cos^230^\circ - \cos^260^\circ$
$\cos(2x - 6) = \cos^2 (90^\circ - 60^\circ ) - \cos^2 60^\circ$
$\cos(2x - 6) = \sin^2 60^\circ - \cos^2 60^\circ$
$\cos (2 x-6)=1-2 \cos ^2 60^{\circ}=1-2\left(\frac{1}{2}\right)^2=1-\frac{1}{2}=\frac{1}{2}$
$\cos(2x - 6) = 1/2$
$\cos(2x - 6) = cos60^\circ$
$(2x - 6) = 60^\circ$
$2x = 66^\circ$
Hence, $x = 33^\circ$
View full question & answer→Question 273 Marks
Find the value of x, if tan $x=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$
Answer$\tan x=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}} $
$ \tan x=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \cdot \frac{1}{\sqrt{3}}} $
$ \tan x=\frac{\frac{3-1}{\sqrt{3}}}{1+1}=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$
Hence, $x = 30^\circ$
View full question & answer→Question 283 Marks
A triangle ABC is right angles at B; find the value of
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B}$
AnswerSince, $ABC$ is a right angled triangle, right angled at $B.$
So, $A + C = 90^\circ$
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B} $
$=\frac{\sec \left(90^{\circ}-C\right) \cdot \cos e c C-\tan \left(90^{\circ}-C\right) \cdot \cot C}{\sin 90^{\circ}}$
$ =\frac{\cos e c C \cdot \cos e c C-\cot C \cdot \cot C}{1} $
$ =1\left(\because \operatorname{cosec}^2 \theta-\cot ^2 \theta=1\right)$
View full question & answer→Question 293 Marks
Evaluate:
$\frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ} \sec 31^{\circ}$
Answer$\frac{\sin 80^{\circ}}{\cos 10^{\circ}}+\sin 59^{\circ} \sec 31^{\circ}$
$ =\frac{\sin \left(90^{\circ}-10^{\circ}\right)}{\cos 10^{\circ}}+\sin \left(90^{\circ}-31^{\circ}\right) \sec 31^{\circ}$
$ =\frac{\cos 10^{\circ}}{\cos 10^{\circ}}+\frac{\cos 31^{\circ}}{\cos 31^{\circ}} $
$=1+1=2$
View full question & answer→Question 303 Marks
Show that:
$\sin A \cos A-\frac{\sin A \cos \left(90^{\circ}-A\right) \cos A}{\sec \left(90^{\circ}-A\right)}-\frac{\cos A \sin \left(90^{\circ}-A\right) \sin A}{\operatorname{cosec}\left(90^{\circ}-A\right)}=0$
Answer$\sin A \cos A-\frac{\sin A \cos \left(90^{\circ}-A\right) \cos A}{\sec \left(90^{\circ}-A\right)}-\frac{\cos A \sin \left(90^{\circ}-A\right) \sin A}{\operatorname{cosec}\left(90^{\circ}-A\right)}$
$=\sin A \cos A-\frac{\sin A \sin A \cos A}{\cos e c A}-\frac{\cos A \cos A \sin A}{\sec A}$
$= \sin A \cos A - \sin^3 A \cos A - \cos^3 A \sin A$
$= \sin A \cos A - \sin A \cos A (\sin^2A + \cos^2A)$
$= \sin A \cos A - \sin A \cos A(1)$
$= 0$
View full question & answer→Question 313 Marks
Show that:
sin 42° sec 48° + cos 42° cos ec48° = 2
Answersin 42° sec 48° + cos 42° cos ec48° = 2
consider sin42° sec48° + cos42° cosec48°
⇒ sin42° sec(90° - 42°) + cos42° cosec(90° - 42°)
⇒ sin42° cosec42° + cos42° sec42°
$\Rightarrow \sin 42^{\circ}, \frac{1}{\sin 42^{\circ}+} \frac{\cos 42^{\circ} 1}{\cos 42^{\circ}}$
⇒ 1+ 1 = 2
View full question & answer→Question 323 Marks
Evaluate:
$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
Answer$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
$\frac{\sin 35^{\circ} \cdot \cos \left(90^{\circ}-35^{\circ}\right)+\cos 35^{\circ} \cdot \sin \left(90^{\circ}-35^{\circ}\right)}{\cos e c^2\left(90^{\circ}-80^{\circ}\right)-\tan ^2 80^{\circ}} $
$ =\frac{\sin 35^{\circ} \cdot \sin 35^{\circ}+\cos 35^{\circ} \cdot \cos 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}} $
$ =\frac{\sin ^2 35^{\circ}+\cos ^2 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}}=\frac{1}{1}=1$
View full question & answer→Question 333 Marks
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : $m^2 - n^2 = a^2 - b^2$
AnswerGiven,
$m = a secA + b tanA and n = a tanA + b secA$
$m^2 - n^2= (a secA + b tanA)^2 - ( a tanA + b secA)^2$
$= a^2 sec^2A + b^2tan^2A + 2ab\ secA\ tanA - (a^2 \tan^2A + b^2 sec^2A + 2ab\ secA\ tanA)$
$= sec^2A (a^2 - b^2) + \tan^2A (b^2 - a^2)$
$= (a^2 - b^2) [sec^2A - \tan^2A]$
$=(a^2 - b^2) [ sincesec^2A - \tan^2A = 1]$
Hence,$m^2 - n^2 = a^2 - b^2$
View full question & answer→Question 343 Marks
If $x \cos A+\sin A=m$ and
$X \sin A-y \cos A=n$, then prove that: $x^2+y^2=m^2+n^2$
Answer$m^2+n^2$
$=(x \cos A+y \sin A)^2+(x \sin A-y \cos A)^2$
$=x^2 \cos ^2 A+y^2 \sin ^2 A+2 x y \sin A \cos A+x^2 \sin ^2 A+y^2 \cos ^2 A-2 x y \sin A \cos A$
$=x^2\left(\cos ^2 A+\sin ^2 A\right)+y^2\left(\cos ^2 A+\sin ^2 A\right)$
$=x^2+y^2$
Hence, $x^2+y^2=m^2+n^2$
View full question & answer→Question 353 Marks
Prove.
$(1 + tanA\ tanB)^2 + (tanA - tanB)^2 = sec^2A sec^2B$
Answer$LHS =(1 + tanA tanB)^2 + (tanA - tanB)^2$
$= 1 + \tan^2A \tan^2B + 2tanA\ tanB+\tan^2A + \tan^2B - 2 tanA tanB$
$= 1 + \tan^2A + \tan^2B + \tan^2A \tan^2B$
$= sec^2A + \tan^2B (1 + \tan^2A)$
$= sec^2A + \tan^2B sec^2A$
$= sec^2A (1 + \tan^2B)$
$= sec^2A sec^2B = RHS$
View full question & answer→Question 363 Marks
Prove.
$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
Answer$\text { LHS }=(\cos e c A-\sin A)(\sec A-\cos A)$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$
$ =\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right) $
$ =\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right) $
$ =\sin A \cos A \\ \text { RHS }=\frac{1}{\tan A+\cot A}$
$=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}} $
$=\frac{\sin A \cos A}{\sin A+\cos 2} A $
$=\sin A \cos A $
$ \text { LHS }=\text { RHS }$
View full question & answer→Question 373 Marks
Prove.
$\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B}=0$
Answer$\text { LHS }=\frac{\sin A-\sin B}{\cos A+\cos B}+\frac{\cos A-\cos B}{\sin A+\sin B} $
$=\frac{\sin ^2 A-\sin ^2 B+\cos ^2 A-\cos ^2 B}{(\cos A+\cos B)(\sin A+\sin B)} $
$=\frac{\left(\sin ^2 A+\cos ^2 A\right)-\left(\sin ^2 B+\cos ^2 B\right)}{(\cos A+\cos B)(\sin A+\sin B)} $
$\frac{1-1}{(\cos A+\cos B)(\sin A+\sin B)} $
$ =0=\text { RHS }$
View full question & answer→Question 383 Marks
Prove.
$\left(\tan A+\frac{1}{\cos A}\right)^2+\left(\tan A-\frac{1}{\cos A}\right)^2=2\left(\frac{1+\sin ^2 A}{1-\sin ^2 A}\right)$
Answer$\text { LHS }=\left(\tan A+\frac{1}{\cos A}\right)^2+\left(\tan A-\frac{1}{\cos A}\right)^2$
$=\left(\frac{\sin A+1}{\cos A}\right)^2+\left(\frac{\sin A-1}{\cos A}\right)^2 $
$ =\frac{\sin ^2 A+1+2 \sin A+\sin ^2 A+1-2 \sin A}{\cos ^2 A}$
$=\frac{2+2 \sin ^2 A}{\cos ^2 A}$
$ =2\left(\frac{1+\sin ^2 A}{1-\sin ^2 A}\right)=\text { RHS }$
View full question & answer→Question 393 Marks
Prove.
$\frac{\tan A}{1-\cot A}+\frac{\cot }{1-\tan A}=\sec A \operatorname{cosec} A+1$
Answer$\text { LHS }=\frac{\tan A}{1-\cot A}+\frac{\cot }{1-\tan A} $
$ =\frac{\tan A}{1-\frac{1}{\tan A}}+\frac{\frac{1}{\tan A}}{1-\tan A}$
$=\frac{\tan ^2 A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)} $
$=\frac{\tan ^3 A-1}{\tan A(1-\tan A)}$
$ =\frac{\left(\tan ^2-1\right)(\tan A+1+\tan A)}{\tan A(\tan A-1)} $
$ =\frac{\sec ^2 A+\tan A}{\tan A}$
$ =\frac{\frac{1}{\cos ^2 A}}{\frac{\sin A}{\cos ^2}+1}=\frac{1}{\sin A \cos A}+1 $
$ =\sec A \operatorname{cosec} A+1=\text { RHS }$
View full question & answer→Question 403 Marks
Prove.
$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$
Answer$\text { LHS }=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A} $
$ =\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}} $
$ \frac{\cos ^2 A}{\cos A-\sin A}+\frac{\sin ^2 A}{\sin A-\cos A}=\frac{\cos ^2 A-\sin ^2 A}{\cos A-\sin A} $
$ =\sin A+\cos A=\text { RHS }$
View full question & answer→Question 413 Marks
Prove the following trigonometric identities.
$sec A (1 − sin A) (sec A + tan A) = 1$
AnswerWe have to prove $\sec \mathrm{A}(1-\sin \mathrm{A})(\sec \mathrm{A}+\tan \mathrm{A})=1$
We know that $\sec ^2 \mathrm{~A}-\tan ^2 \mathrm{~A}-1$
So,
$\sec A(1-\sin A)(\sec A+\tan A)=\{\sec A(1-\sin A)\}(\sec A+\tan A)$
$=(\sec A-\sec A \sin A)(\sec A+\tan A)$
$=\left(\sec A-\frac{1}{\cos A} \sin A\right)(\sec A+\tan A) \ldots\left(\because \sec \theta=\frac{1}{\cos \theta}\right)$
$=\left(\sec A-\frac{\sin A}{\cos A}\right)(\sec A+\tan A) \ldots\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$
$=(\sec A-\tan A)(\sec A+\tan A)$
$=\sec ^2 A-\tan ^2 A$
$=1=\text { R.H.S. }\left(\because \sec ^2 \theta=1 \tan ^2 \theta\right)$
View full question & answer→Question 423 Marks
Prove.
$\frac{\cos \theta \cot \theta}{1+\sin \theta}=\operatorname{cosec} \theta-1$
Answer$\text { LHS }=\frac{\cos \theta \cot \theta}{1+\sin \theta} $
$=\frac{\cos \theta \cot \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta} $
$ =\frac{\cos \theta \cot \theta(1-\sin \theta)}{1-\sin \theta}$
$ =\frac{\cos \theta \frac{\cos \theta}{\sin \theta}(1-\sin \theta)}{\cos ^2 \theta}$
$=\frac{1-\sin \theta}{\sin \theta}$
$=\frac{1}{\sin \theta}-1 $
$=\operatorname{cosec} \theta-1$
View full question & answer→Question 433 Marks
Prove.
$\frac{1+\sin A}{\operatorname{cosec} A-\cot A}-\frac{1-\sin A}{\operatorname{cosec} A+\cot A}=2(1+\cot A)$
Answer$\text { LHS }=\frac{1+\sin A}{\operatorname{cosec} A-\cot A}-\frac{1-\sin A}{\operatorname{cosec} A+\cot A} $
$=\frac{(1+\sin A)(\operatorname{cosec} A+\cot A)-(1-\sin A)(\operatorname{cosec} A-\cot A)}{(\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A)} $
$ =\frac{\operatorname{cosec} A+\cot A+\sin A \operatorname{cosec} A+\sin A \cot A-\operatorname{cosec} A+\cot A+\sin A \operatorname{cosec} A-\sin A \cos A}{\operatorname{cosec}^2 A-\cot ^2 A} $
$ =2 \cot A+2 \sin A \operatorname{cosec} A $
$ =2 \cot A+2 \frac{1}{\operatorname{cosec} A} \times \operatorname{cosec} A $
$=2(\cot A+1) $
Hence proved.
View full question & answer→Question 443 Marks
Prove.$\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1}=\frac{1+\cos A}{\sin A}$
Answer$\text { LHS }=\frac{\cot A+\operatorname{cosec} A-1}{\cot A-\operatorname{cosec} A+1} $
$ =\frac{\cot A+\operatorname{cosec} A-\left(\operatorname{cosec}{ }^2 A-\cot ^2 A\right)}{\cot A-\operatorname{cosec} A+1}\left(\operatorname{cosec}^2 A-\cot ^2 A=1\right) $
$ =\frac{\cot A+\operatorname{cosec} A-((\operatorname{cosec} A-\cot A)(\operatorname{cosec} A+\cot A))}{\cot A-\operatorname{cosec} A+1} $
$ =\frac{\cot A+\operatorname{cosec} A(1-\operatorname{cosec} A+\cot A)}{\cot A-\operatorname{cosec} A+1}$
$ =\cot A +\operatorname{cosec} A $
$ =\frac{\cos A}{\sin A}+\frac{1}{\sin A}$
$=\frac{1+\cos A}{\sin A}$
View full question & answer→Question 453 Marks
Prove.
$\frac{\sin A \tan A}{1-\cos A}=1+\sec A$
Answer$\text { LHS }=\frac{\sin A \tan A}{1-\cos A} $
$ =\frac{\sin A \tan A}{1-\cos A} \times \frac{1+\cos A}{1+\cos A} $
$ =\frac{\sin A \tan A(1+\cos A)}{1-\cos ^2 A} $
$=\frac{\sin A \frac{\sin A}{\cos A}(1+\cos A)}{\sin ^2 A} $
$ =\frac{1+\cos A}{\cos A}$
$ =\frac{1}{\cos A}+\frac{\cos A}{\cos A} $
$ =\sec A+1$
View full question & answer→Question 463 Marks
Prove.
$\frac{\cos A}{1-\sin A}=\sec A+\tan A$
Answer$\text { LHS }=\frac{\cos A}{1-\sin A} $
$\text { RHS }=\sec A+\tan A $
$=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A} $
$ =\frac{1+\sin A}{\cos A}\left(\frac{1-\sin A}{1-\sin A}\right)=\left(\frac{1-\sin ^2 A}{\cos A(1-\sin A)}\right) $
$=\frac{\cos ^2 A}{\cos A(1-\sin A)}=\frac{\cos A}{1-\sin A}=\text { LHS }$
View full question & answer→Question 473 Marks
Prove.
$\frac{\sin A}{1+\cos A}=\operatorname{cosec} A-\cot A$
Answer$\text { LHS }=\frac{\sin A}{1+\cos A} $
$=\frac{\sin A}{1+\cos A} \times \frac{1-\cos A}{1-\cos A} $
$ =\frac{\sin A(1-\cos A)}{1-\cos ^2 A}$
$=\frac{1-\cos A}{\sin A} $
$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$
$=\operatorname{cosec} A-\cot A=\text { RHS }$
View full question & answer→Question 483 Marks
Prove.
$\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A}=\tan A$
Answer$\text { LHS }=\frac{\sin A-2 \sin ^3 A}{2 \cos ^3 A-\cos A} $
$ =\frac{\sin A\left(1-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-1\right)} $
$ =\frac{\sin A\left(\sin ^2 A+\cos ^2 A-2 \sin ^2 A\right)}{\cos A\left(2 \cos ^2 A-\sin ^2 A-\cos ^2 A\right)} $
$=\frac{\sin A\left(\cos ^2 A-\sin ^2 A\right)}{\cos A\left(\cos ^2 A-\sin ^2 A\right)} $
$=\frac{\sin A}{\cos A} $
$=\text { tan A = RHS }$
View full question & answer→Question 493 Marks
Prove.
$\tan ^2 A-\tan ^2 B=\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}$
Answer$\text { LHS }=\tan ^2 A-\tan ^2 B$
$=\frac{\sin ^2 A}{\cos ^2 A}-\frac{\sin ^2 B}{\cos ^2 B} $
$ =\frac{\sin ^2 A \cdot \cos ^2 B-\sin ^2 B \cdot \cos ^2 A}{\cos ^2 A \cdot \cos ^2 B}$
$=\frac{\sin ^2 A\left(1-\sin ^2 B\right)-\sin ^2 B\left(1-\sin ^2 A\right)}{\cos ^2 A \cdot \cos ^2 B} $
$ =\frac{\sin ^2 A-\sin ^2 A \cdot \sin ^2 B-\sin ^2 B+\sin ^2 A \cdot \sin ^2 B}{\cos ^2 A \cdot \cos ^2 B} $
$=\frac{\sin ^2 A-\sin ^2 A \cdot \sin ^2 B-\sin ^2 B+\sin ^2 A \cdot \sin ^2 B}{\cos ^2 A \cdot \cos ^2 B} $
$ =\frac{\sin ^2 A-\sin ^2 B}{\cos ^2 A \cdot \cos ^2 B}=\text { RHS }$
View full question & answer→Question 503 Marks
Prove.
$\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\left(\frac{\cos A}{1+\sin A}\right)^2$
Answer$\text { LHS }=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} $
$ =\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} \times \frac{\operatorname{cosec} A+1}{\operatorname{cosec} A+1}$
$ =\frac{\operatorname{cosec} A-1}{(\operatorname{cosec} A+1)^2} $
$=\frac{\cot ^2 A}{(\operatorname{cosec} A+1)^2} $
$=\frac{\frac{\cos ^2 A}{\sin ^2 A}}{\left(\frac{1}{\sin A}+1\right)^2} $
$=\left(\frac{\cos A}{1+\sin A}\right)^2=\text { RHS }$
View full question & answer→