Question 12 Marks
If A and B are complementary angles, prove that:
$\cot A \cot B-\sin A \cos B-\cos A \sin B=0$
Answer$\text { Since, } A \text { and } B \text { are complementary angles, } A+B=90^{\circ}$
$\cot A \cot B-\sin A \cos B-\cos A \sin B$
$=\cot A \cot \left(90^{\circ}-A\right)-\sin A \cos \left(90^{\circ}-A\right)-\cos A \sin \left(90^{\circ}-A\right)$
$=\cot A \tan A-\sin A \sin A-\cos A \cos A$
$=1-\left(\sin ^2 A+\cos ^2 A\right)$
$=1-1$
$=0$
View full question & answer→Question 22 Marks
Evaluate$\frac{\cos 75^{\circ}}{\sin 15^{\circ}}+\frac{\sin 12^{\circ}}{\cos 78^{\circ}}-\frac{\cos 18^{\circ}}{\sin 72^{\circ}}$
Answer$\frac{\cos 75^{\circ}}{\sin 15^{\circ}}+\frac{\sin 12^{\circ}}{\cos 78^{\circ}}-\frac{\cos 18^{\circ}}{\sin 72^{\circ}} $
$=\frac{\cos \left(90^{\circ}-15^{\circ}\right)}{\sin 15^{\circ}}+\frac{\sin \left(90^{\circ}-78^{\circ}\right)}{\cos 78^{\circ}}-\frac{\cos \left(90^{\circ}-72^{\circ}\right)}{\sin 72^{\circ}} $
$=\frac{\sin 15^{\circ}}{\sin 15^{\circ}}+\frac{\cos 78^{\circ}}{\cos 78^{\circ}}-\frac{\sin 72^{\circ}}{\sin 72^{\circ}} $
$=1+1-1=1$
View full question & answer→Question 32 Marks
Evaluate
3 cos80° cosec10°+ 2 cos59° cosec31°
Answer3 cos80° cosec10°+ 2 cos59° cosec31°
= 3 cos(90° - 10°) cosec10° + 2 cos(90° - 31°) cosec31°
= 3 sin10° cosec10° + 2 sin31° cosec31°
= 3 + 2 = 5
View full question & answer→Question 42 Marks
Evaluate $\frac{3 \sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}$
Answer$\frac{3 \sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\cos e c 58^{\circ}} $
$=\frac{3 \sin \left(90^{\circ}-18^{\circ}\right)}{\cos 18^{\circ}}-\frac{\sec \left(90^{\circ}-58^{\circ}\right)}{\cos e c 58^{\circ}} $
$ =\frac{3 \cos 18^{\circ}}{\cos 18^{\circ}}-\frac{\operatorname{cosec} 58^{\circ}}{\operatorname{cosec} 58^{\circ}} \\ =3-1=2$
View full question & answer→Question 52 Marks
Evaluate
sin27° sin63° - cos63° cos27°
Answersin27° sin63° - cos63° cos27°
= sin(90° - 63°) sin63° - cos63° cos(90° - 63°)
= cos63° sin63° - cos63° sin63°
= 0
View full question & answer→Question 62 Marks
Evaluate
cos40° cosec50° + sin50° sec40°
Answercos40° cosec50° + sin50° sec40°
= cos(90°-50°) cosec50° + sin(90° - 40°) sec40°
= sin50° cosec50° + cos40° sec40°
= 1+ 1 = 2
View full question & answer→Question 72 Marks
Evaluate $\frac{5 \sin 66^{\circ}}{\cos 24^{\circ}}-\frac{2 \cot 85^{\circ}}{\tan 5^{\circ}}$
Answer$\frac{5 \sin 66^{\circ}}{\cos 24^{\circ}}-\frac{2 \cot 85^{\circ}}{\tan 5^{\circ}} $
$=\frac{5 \sin \left(90^{\circ}-24^{\circ}\right)}{\cos 24^{\circ}}-\frac{2 \cot \left(90^{\circ}-5^{\circ}\right)}{\tan 5^{\circ}} $
$=\frac{5 \cos 24^{\circ}}{\cos 24^{\circ}}-\frac{2 \tan 5^{\circ}}{\tan 5^{\circ}} $
$ =5-2=3$
View full question & answer→Question 82 Marks
Evaluate $\sec 26^{\circ} \sin 64^{\circ}+\frac{\cos e c 33^{\circ}}{\sec 57^{\circ}}$
Answer$\sec 26^{\circ} \sin 64^{\circ}+\frac{\cos e c 33^{\circ}}{\sec 57^{\circ}}$
$ =\sec \left(90^{\circ}-64^{\circ}\right) \sin 64^{\circ}+\frac{\operatorname{cosec}\left(90^{\circ}-57^{\circ}\right)}{\sec 57^{\circ}}$
$ =\cos e c 64^{\circ} \sin 64^{\circ}+\frac{\sec 57^{\circ}}{\sec 57^{\circ}} $
$=1+1=2$
View full question & answer→Question 92 Marks
If $sinA + cosA = p$
and $secA + cosecA = q$, then prove that: $q(p^2 - 1) = 2p$
Answer$q(p^2 - 1) = (secA + cosecA) [(sinA + cosA)^2 - 1]$
$= (secA + cosecA) [(\sin^2A + \cos^2A + 2sinA cosA) - 1]$
$= (secA + cosecA) [(1 + 2sinA cosA)-1]$
$= (secA + cosecA) (2sinA cosA)$
$= 2sinA + 2cosA$
$= 2p$
View full question & answer→Question 102 Marks
Find $A$, if $0^\circ \leq A \leq 90^\circ $ and $\cos^2A - cosA = 0$
Answer$\cos^2A - cosA = 0$
$\Rightarrow cosA (cosA - 1) = 0$
$\Rightarrow cosA = 0 or cosA = 1$
We know $\cos 90^\circ = 0$ and cos $0^\circ = 1$
Hence, $A = 90^\circ $ or $0^\circ$
View full question & answer→Question 112 Marks
Find A, if $0^\circ \leq A \leq 90^\circ$ and $4sin^2A - 3 = 0$
Answer$4 \sin ^2 A -3=0$
$\Rightarrow \sin ^2 A=\frac{3}{4}$
$\Rightarrow \sin A=\frac{\sqrt{3}}{2}$
We know $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
Hence, $A=60^{\circ}$
View full question & answer→Question 122 Marks
Find A, if $0^\circ \leq A \leq 90^\circ$ and $2cos^2A - 1 = 0$
Answer$2\cos^2 A - 1 = 0$
$\Rightarrow \cos ^2 A=\frac{1}{2}$
$\Rightarrow \cos A=\frac{1}{\sqrt{2}}$
We know $\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
Hence, $A=45^{\circ}$
View full question & answer→Question 132 Marks
If $4 \cos ^2 A-3=0$ and $\leq A \leq 90^{\circ}$, then prove that:
$\cos 3 \mathrm{~A}=4 \cos ^3 \mathrm{~A}-3 \cos \mathrm{~A}$
Answer$LHS = \cos 3A = \cos 90 = 0$
$RHS = 4 \cos^3 A - 3\cos A$
$= 4 \cos^3 30 - 3\cos 30$
$=4\left(\frac{\sqrt{3}}{2}\right)^3-3\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}=0$
LHS = RHS
View full question & answer→Question 142 Marks
Prove the following identitie:
$1-\frac{\sin ^2 A}{1+\cos A}=\cos A$
Answer$1-\frac{\sin ^2 A}{1+\cos A}$
$=\frac{1+\cos A-\sin ^2 A}{1+\cos A} $
$ =\frac{\cos A+\cos ^2 A}{1+\cos A} $
$=\frac{\cos A(1+\cos A)}{1+\cos A}$
$ =\cos A$
View full question & answer→Question 152 Marks
Prove that
(sinA + cosA) (secA + cosecA) = 2 + secA cosecA
Answer$(\sin A +\cos A )(\sec A +\operatorname{cosec} A )$
$=\frac{\sin A}{\cos A}+1+1+\frac{\cos A}{\sin A}$
$=2+\frac{\cos ^2 A+\sin ^2 A}{\sin A \cos A}$
$=2+\frac{1}{\sin A \cos A}$
$=2+\sec A \operatorname{cosec} A$
View full question & answer→Question 162 Marks
Prove that
$(\sin A-\cos A)(1+\tan A+\cot A)=\frac{\sec A}{\operatorname{cosec} 2}-\frac{\operatorname{cosec} A}{\sec ^2 A}$
Answer$(\sin A-\cos A)(1+\tan A+\cot A) $
$ =\sin A+\frac{\sin ^2 A}{\cos A}+\cos A-\cos A-\sin A-\frac{\cos ^2 A}{\sin A} $
$ =\frac{\sin ^2 A}{\cos A}-\frac{\cos ^2 A}{\sin A} $
$=\frac{\sec A}{\operatorname{cosec} 2}-\frac{\operatorname{cosec} A}{\sec ^2 A}$
View full question & answer→Question 172 Marks
Prove that
cosA (1 + cotA) + sinA (1 + tanA = secA + cosecA
Answer$\cos A (1+\cot A )+\sin A (1+\tan A )$
$ =\cos A+\frac{\cos ^2 A}{\sin A}+\sin A+\frac{\sin ^2 A}{\cos A} $
$=\sin A+\frac{\cos ^2 A}{\sin A}+\cos A+\frac{\sin ^2 A}{\cos A} $
$ =\left(\frac{\cos ^2 A+\sin ^2 A}{\sin A}\right)+\left(\frac{\cos ^2 A+\sin ^2 A}{\cos A}\right) $
$ =\frac{1}{\sin A}+\frac{1}{\cos A} $
$ =\operatorname{cosec} A+\sec A$
View full question & answer→Question 182 Marks
Prove that
$\frac{\cos A}{1+\sin A}=\sec A-\tan A$
Answer$\begin{aligned} & \frac{\cos A}{1+\sin A} \\ = & \frac{\cos A}{1+\sin A} \times \frac{1-\sin A}{1-\sin A} \\ = & \frac{\cos A(1-\sin A)}{1-\sin ^2 A} \\ = & \frac{\cos A(1-\sin A)}{\cos ^2 A} \\ = & \frac{1-\sin A}{\cos A} \\ = & \sec A-\tan A\end{aligned}$
View full question & answer→Question 192 Marks
Prove that
$\frac{\cot ^2 A}{\operatorname{cosec} A-1}-1=\operatorname{cosec} A$
Answer$\frac{\cot ^2 A}{\operatorname{cosec} A-1}-1 $
$=\frac{\cot ^2 A-\operatorname{cosec} A+1}{\operatorname{cosec} A-1}$
$ =\frac{-\operatorname{cosec} A+\operatorname{cosec}{ }^2 A}{\operatorname{cosec} A-1}$
$=\frac{\operatorname{cosec} A(\operatorname{cosec} A-1)}{\operatorname{cosec} A-1}$
$ =\operatorname{cosec} A$
View full question & answer→Question 202 Marks
Use tables to find the acute angle θ, if the value of tan θ is 0.7391
AnswerFrom the tables, it is clear that tan 36° 24’ = 0.7373
tan θ − tan 36° 24’ = 0.7391 − 0.7373 = 0.0018
From the tables, diff of 4’ = 0.0018
Hence, θ = 36° 24’ + 4’ = 36° 28’
View full question & answer→Question 212 Marks
Use tables to find the acute angle θ, if the value of tan θ is 0.4741
AnswerFrom the tables, it is clear that tan 25° 18’ = 0.4727
tan θ − tan 25° 18’ = 0.4741 − 0.4727 = 0.0014
From the tables, diff of 4’ = 0.0014
Hence, θ = 25° 18’ + 4’ = 25° 22’
View full question & answer→Question 222 Marks
Use tables to find the acute angle θ, if the value of cos θ is 0.6885
AnswerFrom the tables, it is clear that cos 46° 30’ = 0.6884
cos q − cos 46° 30’ = 0.6885 − 0.6884 = 0.0001
From the tables, diff of 1’ = 0.0002
Hence, θ = 46° 30’ − 1’ = 46° 29’
View full question & answer→Question 232 Marks
Use tables to find the acute angle θ, if the value of cos θ is 0.9574
AnswerFrom the tables, it is clear that cos 16° 48’ = 0.9573
cos θ − cos 16° 48’ = 0.9574 − 0.9573 = 0.0001
From the tables, diff of 1’ = 0.0001
Hence, θ = 16° 48’ − 1’ = 16° 47’
View full question & answer→Question 242 Marks
Use tables to find the acute angle θ, if the value of sin θ is 0.6525
AnswerFrom the tables, it is clear that sin 40° 42' = 0.6521
sin θ − sin 40° 42' = 0.6525 −; 0.6521 = 0.0004
From the tables, diff of 2' = 0.0004
Hence, θ = 40° 42' + 2' = 40° 44
View full question & answer→Question 252 Marks
Prove that:
$\frac{\sin \theta \sin \left(90^{\circ}-\theta\right)}{\cot \left(90^{\circ}-\theta\right)}=1-\sin ^2 \theta$
Answer$\text { LHS }=\frac{\sin \theta \sin \left(90^{\circ}-\theta\right)}{\cot \left(90^{\circ}-\theta\right)} $
$=\frac{\sin \theta \cos \theta}{\tan \theta} $
$=\frac{\sin \theta \cos \theta}{\frac{\sin \theta}{\cos \theta}}$
$ \cos ^2 \theta=1-\sin ^2 \theta$
View full question & answer→Question 262 Marks
Prove that:
$\frac{\cos \left(90^{\circ}-\theta\right) \cos \theta}{\cot \theta}=1-\cos ^2 \theta$
Answer$\text { LHS }=\frac{\cos \left(90^{\circ}-\theta\right) \cos \theta}{\cot \theta} $
$ =\frac{\sin \theta \cos \theta}{\frac{\cos \theta}{\sin \theta}}=\sin ^2 \theta $
$ =1-\cos ^2 \theta$
View full question & answer→Question 272 Marks
find the value of angle $A,$ where $0^\circ \leq A \leq 90^\circ.$
$\cos(90^\circ - A).\sec 77^\circ = 1$
Answer$\cos \left(90^{\circ}- A \right) \cdot \sec 77^{\circ}=1$
$\sin A \cdot \frac{1}{\cos 77^{\circ}=} 1$
$\sin A=\cos 77^{\circ}=\cos \left(90^{\circ}-13^{\circ}\right)=\sin 13^{\circ}$
$A=13^{\circ}$
View full question & answer→Question 282 Marks
find the value of angle $A,$ where $0^\circ \leq A \leq 90^\circ.$
$\sin(90^\circ - 3A).\cos ec 42^\circ = 1$
Answer$\sin \left(90^{\circ}-3 A\right) \cdot \operatorname{cosec} 42^{\circ}=1$
$\cos 3 A \frac{1}{\sin 42^{\circ}}=1$
$\cos 3 A=\sin 42^{\circ}=\sin \left(90^{\circ}-48^{\circ}\right)=\cos 48^{\circ}$
$3 A=48^{\circ}$
$A=16^{\circ}$
View full question & answer→Question 292 Marks
Find the value of x, if $\sin 3x = 2\sin 30^\circ \cos 30^\circ$
Answer$\sin 3 x=2 \sin 30^{\circ} \cos 30^{\circ}$
$\sin 3 x=2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
$\sin 3 x=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$
$3 x=60^{\circ}$
Hence, $x=20^{\circ}$
View full question & answer→Question 302 Marks
Find the value of x, if $\sin 2 x=2 \sin 45^{\circ} \cos 45^{\circ}$
Answer$\sin 2 x=2 \sin 45^{\circ} \cos 45^{\circ}$
$\sin 2 x=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$
$\sin 2x = 1 = \sin 90^\circ$
$2x = 90^\circ$
Hence, $x = 45^\circ$
View full question & answer→Question 312 Marks
Find the value of x, if $\cos x=\cos 60^{\circ} \cos 30^{\circ}-\sin 60^{\circ} \sin 30^{\circ}$
Answer$\cos x=\cos 60^{\circ} \cos 30^{\circ}-\sin 60^{\circ} \sin 30^{\circ}$
$\cos x=\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
$\cos x=0=\cos 90^{\circ}$
$\text { Hence, } x=90^{\circ}$
View full question & answer→Question 322 Marks
Find the value of x, if $\sin x=\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$
Answer$\sin x=\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$
$\sin x=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
$\sin x=\frac{3}{4}+\frac{1}{4}=1=\sin 90^{\circ}$
Hence, $x=90^{\circ}$
View full question & answer→Question 332 Marks
Find the value of x, if $\sin x=\sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ}$
Answer$\sin x=\sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ}$
$\sin x=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
$\sin x=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}=\sin 30^{\circ}$
Hence, $x=30^{\circ}$
View full question & answer→Question 342 Marks
Evaluate:
$\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}$
Answer$\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ} $
$ =\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}+\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}-\left(\frac{1}{2}\right)^2$
$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}-2$
$=1+1-2=0$
View full question & answer→Question 352 Marks
Evaluate:
$\frac{\cot ^2 41^{\circ}}{\tan ^2 49^{\circ}}-2 \frac{\sin ^2 75^{\circ}}{\cos ^2 15^{\circ}}$
Answer$\frac{\cot ^2 41^{\circ}}{\tan ^2 49^{\circ}}-2 \frac{\sin ^2 75^{\circ}}{\cos ^2 15^{\circ}} $
$ =\frac{\left(\cot \left(90^{\circ}-49^{\circ}\right)\right)^2}{\tan ^2 49^{\circ}}-2 \frac{\left(\sin \left(90^{\circ}-15^{\circ}\right)\right)^2}{\cos ^2 15^{\circ}} $
$ =\frac{\tan ^2 49^{\circ}}{\tan ^2 49^{\circ}}-2 \frac{\cos ^2 15^{\circ}}{\cos ^2 15^{\circ}} $
$ =1-2=-1$
View full question & answer→Question 362 Marks
Evaluate:
$2 \frac{\tan 57^{\circ}}{\cot 33^{\circ}}-\frac{\cot 70^{\circ}}{\tan 20^{\circ}}-\sqrt{2} \cos 45^{\circ}$
Answer$2 \frac{\tan 57^{\circ}}{\cot 33^{\circ}}-\frac{\cot 70^{\circ}}{\tan 20^{\circ}}-\sqrt{2} \cos 45^{\circ} $
$ 2 \frac{\tan \left(90^{\circ}-33^{\circ}\right)}{\cot 33^{\circ}}-\frac{\cot \left(90^{\circ}-20^{\circ}\right)}{\tan 20^{\circ}}-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) $
$2 \frac{\cot 33^{\circ}}{\cot 33^{\circ}}-\frac{\tan 20^{\circ}}{\tan 20^{\circ}}-1$
$=2-1-1 $
$ =0$
View full question & answer→Question 372 Marks
Evaluate:
cosec(65° + A) - sec(25° - A)
Answercosec(65° + A) - sec(25° - A)
= cosec[90° - (25° - A)] - sec(25° - A)
= sec(25° - A) - sec(25° - A)
= 0
View full question & answer→Question 382 Marks
Evaluate:
3cos80° cosec10° + 2 cos59° cosec31°
Answer3cos80° cosec10° + 2 cos59° cosec31°
= 3cos(90° - 10°) cosec10° + 2cos(90° - 31°) cosec31°
= 3sin10° cosec10° + 2sin31° cosec31°
= 3 + 2 = 5
View full question & answer→Question 392 Marks
Evaluate:
$3 \frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}$
Answer$3 \frac{\sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\cos e c 58^{\circ}} $
$ =3 \frac{\sin \left(90^{\circ}-18^{\circ}\right)}{\cos 18^{\circ}}-\frac{\sec \left(90^{\circ}-58^{\circ}\right)}{\cos e c 58^{\circ}}$
$ =3 \frac{\cos 18^{\circ}}{\cos 18^{\circ}}-\frac{\operatorname{cosec} 58^{\circ}}{\operatorname{cosec} 58^{\circ}}=3-1=2$
View full question & answer→Question 402 Marks
For triangle ABC, show that:
$\frac{\tan (B+C)}{2}=\frac{\cot A}{2}$
AnswerWe know that for a triangle $\triangle A B C$
$\angle A +\angle B +\angle C =180^{\circ}$
$\angle B +\angle C =180^{\circ}-\angle A $
$ \Rightarrow \frac{\angle B +\angle C }{2}=90-\frac{\angle A }{2} $
$ \Rightarrow \tan \left(\frac{ B + C }{2}\right)=\tan \left(90^{\circ}-\frac{ A }{2}\right)$
$=\cot \left(\frac{ A }{2}\right)$
View full question & answer→Question 412 Marks
For triangle ABC, show that:
$\frac{\sin (A+B)}{2}=\frac{\cos C}{2}$
AnswerWe know that for a triangle $\triangle ABC$
$\angle A +\angle B +\angle C = 180^\circ$
$\frac{\angle B+\angle A}{2}=90-\frac{\angle C}{2}$
$\sin \left(\frac{A+B}{2}\right)=\sin \left(90^{\circ}-\frac{C}{2}\right) $
$=\cos \left(\frac{C}{2}\right)$
View full question & answer→Question 422 Marks
Show that:
$\frac{\sin A}{\sin \left(90^{\circ}-A\right)}+\frac{\cos A}{\cos \left(90^{\circ}-A\right)}=\sec A \operatorname{cosec} A$
Answer$\frac{\sin A}{\sin \left(90^{\circ}-A\right)}+\frac{\cos A}{\cos \left(90^{\circ}-A\right)}$
$ \frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} $
$=\frac{\sin ^2 A+\cos ^2 A}{\cos A \sin A} $
$ =\frac{1}{\cos A \sin A} $
$=\sec A \operatorname{cosec} A$
View full question & answer→Question 432 Marks
Show that:
$\frac{\sin 26^{\circ}}{\sec 64^{\circ}+} \frac{\cos 26^{\circ}}{\operatorname{cosec} 64^{\circ}}=1$
Answer$\frac{\sin 26^{\circ}}{\sec 64^{\circ}+} \frac{\cos 26^{\circ}}{\cos e c 64^{\circ}} $
$ =\frac{\sin 26^{\circ}}{\sec \left(90^{\circ}-26^{\circ}\right)}+\frac{\cos 26^{\circ}}{\operatorname{cosec}\left(90^{\circ}-26^{\circ}\right)} $
$=\frac{\sin 26^{\circ}}{\operatorname{cosec} 26^{\circ}}+\frac{\cos 26^{\circ}}{\sec 26^{\circ}} $
$ =\sin ^2 26^{\circ}+\cos ^2 26^{\circ} $
$ =1$
View full question & answer→Question 442 Marks
Show that:
tan10° tan15° tan75° tan80° = 1
Answertan10° tan15° tan75° tan80° = 1
= tan(90°- 80°) tan(90° -75°) tan75° tan80°
= cot80° cot75° tan75° tan80°
= 1[As tanθ, cotθ = 1]
View full question & answer→Question 452 Marks
If $x = r \cos A \cos B, y = r$ cos $A$ sin $B$ and $Z = r$ sin $A,$ show that:
$x^2 + y^2 + z^2 = r^2$
Answer$LHS = (r cosA cosB)^2 + (r cosA sinB)^2 + (r sinA)^2$
$= r^2 \cos^2A \cos^2B + r^2 \cos^2A \sin^2B + r^2 \sin^2A$
$= r^2 \cos^2A (\cos^2B + \sin^2B) + r^2 \sin^2A$
$= r^2 (\cos^2A + \sin^2A) = r^2 = RHS$
View full question & answer→Question 462 Marks
If $x = r$ sin $A \cos B, y = r \sin A \sin B$ and $z = r \cos A$, then prove that: $x^2 + y^2 + z^2 = r^2$
Answer$LHS = (r sinA cosB)^2 + (r sinA sinB)^2 + (r cosA)^2$
$= r^2sin^2A \cos^2B + r^2sin^2A \sin^2B + r^2 \cos^2A$
$= r^2sin^2A (\cos^2B +\sin^2B) +r^2cos^2A$
$=r^2(\sin^2A +\cos^2A) =r^2= RHS$
View full question & answer→Question 472 Marks
Prove.
$2 \sin^2A + \cos^4A = 1 + \sin^4A$
Answer$LHS =2 \sin^2A + \cos^4A$
$= 2 \sin^2A + (1 - \sin^2A)^2$
$= 2 \sin^2A + 1 + \sin^4A - 2 \sin^2A$
$= 1 + \sin^4A = RHS$
View full question & answer→Question 482 Marks
Prove.
$\operatorname{cosec} A (1+\cos A )(\operatorname{cosec} A-\cot A ) =1$
Answer$\text { LHS }=\operatorname{cosec} A (1+\cos A )(\operatorname{cosec} A-\cot A )$
$=\frac{1}{\sin A}(1+\cos A)\left(\frac{1}{\sin A}-\frac{\cos A}{\sin A}\right)$
$=\frac{(1+\cos A)}{\sin A}\left(\frac{1-\cos A}{\sin A}\right)$
$=\frac{\left(1-\cos ^2 A\right)}{\sin ^2 A}\left(\frac{\sin ^2 A}{\sin ^2 A}\right)=1=\text { RHS }$
View full question & answer→Question 492 Marks
Prove.
$cosec^4A - cosec^2 A = \cot^4 A + \cot^2 A$
Answer$LHS =cosec^4A - cosec^2 A$
$= cosec^2 A (cosec^2A- 1)$
$RHS = \cot^4A + \cot^2 A$
$= \cot^2 A (\cot^2 A + 1)$
$= (cosec^2 A - 1) cosec^2 A$
Thus, $LHS = RHS$
View full question & answer→Question 502 Marks
Prove.
$(1-\tan A)^2+(1+\tan A)^2=2 \sec ^2 A$
Answer$\text { LHS }=(1-\tan A)^2+(1+\tan A)^2$
$=\left(1+\tan ^2 A-2 \tan A\right)+\left(1+\tan ^2 A+2 \tan A\right)$
$=2\left(1+\tan ^2 A\right) $
$=2 \sec ^2 A=\text { RHS }$
View full question & answer→