Maths Class X Question Bank

Question 1 of 31

1. Question
1 point(s)
$\sin(45^o+\theta)-\cos(45^o-\theta) $ is equal to:
- $2\ cosec\ \theta $
- $0 $
- $\sin\theta $
- $1 $
Correct

Incorrect

Hint

$\sin(45+\theta)-\cos(45-\theta)\\ \\ =\sin(45+\theta)=\cos[90-(45+\theta)]=\cos(45-\theta)\\ \\ \cos(45-\theta)-\cos(45-\theta)=0$
Question 2 of 31

2. Question
1 point(s)
$9\sec^2\theta-9\tan^2\theta $ is equal to:
- 1
- 9
- 8
- 0
Correct

Incorrect

Hint

$9\sec^2\theta-9\tan^2\theta=9(sec^2\theta-\tan^2\theta)\\ \\ =9\times1=9$
Question 3 of 31

3. Question
1 point(s)
If $\sin A=\dfrac{8}{17} $ and $A $ is acute, then $\cot A $ is equal to:
- $\dfrac{15}{8} $
- $\dfrac{15}{17} $
- $\dfrac{8}{15} $
- $\dfrac{17}{8} $
Correct

Incorrect

Hint

$\dfrac{p}{h}=\dfrac{8}{17}\Rightarrow b=\sqrt{289-64}=\sqrt{225}=15\\ \\ \cot A=\dfrac{b}{p}=\dfrac{15}{8}$
Question 4 of 31

4. Question
1 point(s)
$(cosec^2\ 72^o-\tan^2\ 18^o) $ is equal to:
- $0 $
- $1 $
- $\dfrac{3}{2} $
- None of these
Correct

Incorrect

Hint

$(cosec^2\ 72-\tan^218)=cosec^2\ (90-18)-\tan^218\\ \\ =\sec^218-\tan^218\\ \\ =1$
Question 5 of 31

5. Question
1 point(s)
If $x=\sec\theta+\tan\theta $, then $\tan\theta $ is equal to:
- $\dfrac{x^2+1}{x} $
- $\dfrac{x^2-1}{x} $
- $\dfrac{x^2+14}{2x} $
- $\dfrac{x^2-1}{2x} $
Correct

Incorrect

Hint

$x=\sec\theta+\tan\theta\\ \\ \sec\theta=x\tan\theta\\ \\ \Rightarrow sec^2\theta=x^2-2x\tan\theta+\tan^2\theta\\ \\ \Rightarrow x^2-2x\tan\theta-1=0\Rightarrow \tan\theta=\dfrac{x^2-x}{2x}$
Question 6 of 31

6. Question
1 point(s)
$\tan^2\theta\sin^2\theta $ is equal to:
- $\tan^2\theta-\sin^2\theta $
- $\tan^2\theta+\sin^2\theta $
- $\dfrac{\tan^2\theta}{\sin^2\theta} $
- None of these
Correct

Incorrect

Hint

$\tan^2\theta\sin^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta}\times \sin^2\theta=\dfrac{\sin^2\theta(1-\cos^2\theta)}{\cos^2\theta}\\ \\ =\dfrac{\sin^2\theta-\sin^2\theta\cos^2\theta}{\cos^2\theta}\\ \\ =\tan^2\theta-\sin^2\theta$
Question 7 of 31

7. Question
1 point(s)
If $\cos\theta-\sin\theta=1 $, then the value of $\cos\theta+\sin\theta $ is equal to:
- $\pm4 $
- $\pm3 $
- $\pm2 $
- $\pm1 $
Correct

Incorrect

Hint

$\cos\theta-\sin\theta=1\Rightarrow \cos\theta+\sin\theta=\pm1$
Question 8 of 31

8. Question
1 point(s)
$\dfrac{1+\tan^2\theta}{1+\cot^2\theta} $ is equal to:
- $\sec^2\theta $
- $-1 $
- $\cot^2\theta $
- $\tan^2\theta $
Correct

Incorrect

Hint

$\dfrac{1+\tan^2\theta}{1+\cot^2\theta}=\dfrac{1+\dfrac{\sin^2\theta}{\cos^2\theta}}{1+\dfrac{\cos^2\theta}{\sin^2\theta}}=\dfrac{\dfrac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}{\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}}\\ \\ =\dfrac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$
Question 9 of 31

9. Question
1 point(s)
$(\sec^210^o-\cot^280^o) $ is equal to
- $1 $
- $0 $
- $2 $
- $\dfrac{1}{2} $
Correct

Incorrect

Hint

$\sec^210-\cot^280\\ \\ =\sec^2(90-80)-\cot^280\\ \\ =cosec^280-\cot^280\\ \\ =\dfrac{1}{\sin^280}-\dfrac{\cos^280}{\sin^280}\\ \\ =\dfrac{1-\cos^280}{\sin^280}=\dfrac{\sin^280}{\sin^280}=1$
Question 10 of 31

10. Question
1 point(s)
The value of $\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}} $ is:
- $\cot\theta-cosec\ \theta $
- $cosec\ \theta+\cot\theta $
- $cosec\ \theta+\cot^2\theta $
- $\cot\theta+cosec^2\theta $
Correct

Incorrect

Hint

$\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}}=\sqrt{\dfrac{(1+\cos\theta)(1+\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}}\\ \\ =\sqrt{\dfrac{(1+\cos\theta)^2}{1^2-\cos^2\theta}}=\dfrac{1+\cos\theta}{\sin\theta}\\ \\ =cosec\theta+\cot\theta$
Question 11 of 31

11. Question
1 point(s)
$\dfrac{\sin\theta}{1+\cos\theta} $ is equal to:
- $\dfrac{1+\cos\theta}{\sin\theta} $
- $\dfrac{1+\cot\theta}{\sin\theta} $
- $\dfrac{1-\cos\theta}{\sin\theta} $
- $\dfrac{1-\sin\theta}{\cos\theta} $
Correct

Incorrect

Hint

$\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)}=\dfrac{\sin\theta(1-\cos\theta)}{\sin^2\theta}\\ \\ =\dfrac{1-\cos\theta}{\sin\theta}$
Question 12 of 31

12. Question
1 point(s)
If $x=a\cos\alpha $ and $y=b\sin\alpha $, then $b^2x^2+a^2y^2 $ is equal to
- $a^2b^2 $
- $ab $
- $a^4b^4 $
- $a^2+b^2 $
Correct

Incorrect

Hint

$x=a\cos\alpha$ and $y=b\sin\alpha$
$b^2x^2+a^2y^2=b^2a^2\cos^2\alpha+a^2b^2\sin^2\alpha\\ \\ =a^2b^2(\cos^2\alpha+\sin^2\alpha)\\ \\ =a^2b^2$
Question 13 of 31

13. Question
1 point(s)
$\sqrt{(1+\sin\theta)(1-\sin\theta)} $ is equal to:
- $\sin\theta $
- $\sin^2\theta $
- $\cos^2\theta $
- $\cos\theta $
Correct

Incorrect

Hint

$\sqrt{(1+\sin\theta)(1-\sin\theta)}=\sqrt{1-\sin^2\theta}\\ \\ =\sqrt{\cos^2\theta}=\cos\theta$
Question 14 of 31

14. Question
1 point(s)
The value of expression $\left[\dfrac{\sin^222^o+\sin^268^o}{\cos^222^o+\cos^268^o}+\sin^263^o+\cos63^o\sin27^o\right] $ is:
- 2
- 1
- 0
- None of these
Correct

Incorrect

Hint

$\dfrac{\sin^222+\sin^268}{\cos^222+\cos^268}+\sin^263+\cos63\sin27\\ \\ =\dfrac{\sin^222+\cos^222}{\sin^268+\cos^268}+\sin^263+\cos63\ \cos63\\ \\ =\dfrac{1}{1}+\sin^263+\cos^263=1+1=2$
Question 15 of 31

15. Question
1 point(s)
If $\cos9\alpha=\sin\alpha $ and $9\alpha<90^o $, then the value of $5\alpha $ is:
- $0 $
- $1 $
- $\sqrt3 $
- None of these
Correct

Incorrect

Hint

$\cos9\alpha=\sin\alpha\\ \\ \cos9\alpha=\cos(90-\alpha)\\ \\ 9\alpha=90-\alpha\Rightarrow 10\alpha=90\Rightarrow \alpha=9\\ \\ \tan5\alpha=\tan5\times9=\tan45=1$
Question 16 of 31

16. Question
1 point(s)
In the given figure, $\angle ACB=90^o,\angle BDC=90^o,CD=4\ cm,\ BD=3\ cm,\ AC=12\ cm,\cos A-\sin A $ is equal to:
- $\dfrac{5}{12} $
- $\dfrac{5}{13} $
- $\dfrac{7}{12} $
- $\dfrac{7}{13} $
Correct

Incorrect

Hint

$BC=\sqrt{16+9}=\sqrt{25}=5\\ \\ AB=\sqrt{144+25}=\sqrt{169}=13\\ \\ \cos A-\sin A=\dfrac{12}{13}-\dfrac{5}{13}=\dfrac{7}{13}$
Question 17 of 31

17. Question
1 point(s)
If $\cot A=\dfrac{12}{5} $, then the value of $(\sin A+\cos A)\ x\ cosec\ A $ is:
- $\dfrac{13}{5} $
- $\dfrac{17}{5} $
- $\dfrac{14}{5} $
- $1 $
Correct

Incorrect

Hint

$\dfrac{b}{P}=\dfrac{12}{5}\Rightarrow h=\sqrt{144+25}=\sqrt{169}=13\\ \\ (\sin A+\cos A)\times cosec A\\ \\ =\left(\dfrac{5}{13}+\dfrac{12}{13}\right)\times \dfrac{13}{5}=\dfrac{17}{13}\times \dfrac{13}{5}=\dfrac{17}{5}$
Question 18 of 31

18. Question
1 point(s)
$\cos1^o,\cos2^o,\cos3^o,…..\cos180^o $ is equal to:
- 1
- 1
- 1/2
- 0
Correct

Incorrect

Hint

$\cos1^o.\cos2^o.\cos3^o….\cos180^o\\ \\ =\cos1^o\ \cos2^o\ \cos3^o…..\cos90….\cos180\\ \\ =\cos1^o\cos2^o\cos3\ \times 0\times…\cos180=0$
Question 19 of 31

19. Question
1 point(s)
$5\ cosec^2\theta-5\cot^2\theta $ is equal to:
- 5
- 1
- 0
- -5
Correct

Incorrect

Hint

$5\ cosec^2\theta-5\cot^2\theta\\ \\ =5(cosec^2\theta-\cot^2\theta)\\ \\ =5\times1=5$
Question 20 of 31

20. Question
1 point(s)
If $\sin\theta=\cos\theta $, then value of $\theta $ is:
- $0^o $
- $45^o $
- $30^o $
- $90^o $
Correct

Incorrect

Hint

$\sin\theta=\cos\theta\\ \\ \dfrac{\sin\theta}{\cos\theta}=1\Rightarrow \tan\theta=1\Rightarrow \theta=45$
Question 21 of 31

21. Question
1 point(s)
$9\sec^2\theta-9\tan^2\theta $ is equal to:
- 1
- -1
- 9
- -9
Correct

Incorrect

Hint

$9\sec^2\theta-9\tan^2\theta=9(\sec^2\theta-\tan^2\theta)\\ \\ =9\times1=9$
Question 22 of 31

22. Question
1 point(s)
If $\sin\theta+\sin^2\theta=1 $, the value of $(\cos^2\theta+\cos^4\theta) $ is
- 3
- 2
- 1
- 0
Correct

Incorrect

Hint

$\sin\theta+\sin^2\theta=1\ \Rightarrow 1-\sin^2\theta=\sin\theta\\ \\ \cos^2\theta+\cos^4\theta=(1-\sin^2\theta)+(1-\sin^2\theta)^2\\ \\ =\sin\theta+\sin^2\theta=1$
Question 23 of 31

23. Question
1 point(s)
In the figure, if D is the mid-point of BC, the value of $\dfrac{\cot y^0}{\cot x^0} $ is:
- $2 $
- $\dfrac{1}{4} $
- $\dfrac{1}{3} $
- $\dfrac{1}{2} $
Correct

Incorrect

Hint

$\dfrac{\cot y}{\cot x}=\dfrac{\dfrac{AC}{BC}}{\dfrac{AC}{CD}}=\dfrac{BC}{CD}=\dfrac{BD+CD}{CD}\\ \\ CD=BD\ \text{so}\ \dfrac{CD+CD}{CD}=\dfrac{2}{1}$
Question 24 of 31

24. Question
1 point(s)
If $cosec\ \theta=\dfrac{3}{2} $, then $2\ (cosec^2\theta+\cot^2\theta) $
- 3
- 7
- 9
- 5
Correct

Incorrect

Hint

$cosec\ \theta=\dfrac{3}{2}\\ \\ 2(cosec^2\theta+cosec^2\theta-1)\\ \\ =2(2\ cosec^2\theta-1)=2\left(2\times\dfrac{9}{4}-1\right)\\ \\ =2\times\dfrac{7}{2}=7$
Question 25 of 31

25. Question
1 point(s)
In the figure, if PS = 13 cm, the value of tan a is equal to :
- $\dfrac{4}{3} $
- $\dfrac{14}{3} $
- $\dfrac{5}{3} $
- $\dfrac{13}{3} $
Correct

Incorrect

Hint

$ST=PS-TP=14-5=9\ cm\\ \\ PR^2=PQ^2+QR^2\Rightarrow 13^2=PQ^2+5^2\\ \\ \Rightarrow PQ^2=144=12\ cm\\ \\ PQ=TR=12\ cm\ \tan\ \alpha=\dfrac{12}{9}=\dfrac{4}{3}$
Question 26 of 31

26. Question
1 point(s)
If $x=3\sec^2\theta-1,\ y=\tan^2\theta-2 $, then $x-3y $ is equal to:
- 3
- 4
- 8
- 5
Correct

Incorrect

Hint

$(3\sec^2\theta-1)-3(\tan^2\theta-2)\\ \\ 3\sec^2\theta-1-3\tan^2\theta+6\\ \\ 3(\sec^2\theta-\tan^2\theta)+5\\ \\ =3\times1+5=3+5=8$
Question 27 of 31

27. Question
1 point(s)
$(\sec A+\tan A)\ (1-\sin A) $ is equal to:
- sec A
- tan A
- sin A
- cos A
Correct

Incorrect

Hint

$(\sec A+\tan A)(1-\sin A)\\ \\ =\left(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right)\ (1-\sin A)\\ \\ =\dfrac{(1+\sin A)(1-\sin A)}{\cos A}=\dfrac{1-\sin^2A}{\cos A}\\ \\ =\dfrac{\cos^2A}{\cos A}=\cos A$
Question 28 of 31

28. Question
1 point(s)
If $\sec\theta-\tan\theta=\dfrac{1}{3} $, the value of $(\sec\theta+\tan\theta) $ is
- 1
- 2
- 3
- 4
Correct

Incorrect

Hint

$\sec^2\theta-\tan^2\theta=1\\ \\ (\sec \theta-\tan\theta)\ (\sec\theta+\tan\theta)=1\\ \\ \sec\theta+\tan\theta=3$
Question 29 of 31

29. Question
1 point(s)
The value of $\dfrac{\cot45^o}{\sin30^o+\cos60^o} $ is equal to:
- $1 $
- $\dfrac{1}{\sqrt2} $
- $\dfrac{2}{3} $
- $\dfrac{1}{2} $
Correct

Incorrect

Hint

$\dfrac{\cot45}{\sin30+\cos60}=\dfrac{1}{\dfrac{1}{2}+\dfrac{1}{2}}=\dfrac{1}{2}=1$
Question 30 of 31

30. Question
1 point(s)
If $\cos2\theta=\dfrac{\sqrt3}{2};0<\theta<20^o $. Then the value of $\theta $ is:
- $15^o $
- $10^o $
- $0^o $
- $12^o $
Correct

Incorrect

Hint

$\cos2\theta=\dfrac{\sqrt3}{2}=\cos30\\ \\ 2\theta=30\\ \\ \theta=15$
Question 31 of 31

31. Question
1 point(s)
$\Delta $ABC is a right angled at A, the value of tan B x tan C is :
- 0
- 1
- -1
- None of these
Correct

Incorrect

Hint

$\tan B\times \tan C\\ \\ =\tan B\times \tan(90-B)\\ \\ \tan B\times \cot B\\ \\ =\tan B\times \dfrac{1}{\tan B}=1$