Call me

Have a Query? We will call you right away.

+91

E.g: 9876543210, 01112345678

We will give you a call shortly, Thank You

Office hours: 9:00 am to 9:00 pm IST (7 days a week)

What are you looking for?

Syllabus

30. Using the measurements given in figure alongside,

(a) find the values of :

(1) sin $\varphi $ (ii) tan $\theta $.

(b) write an expression for AD in terms of $\theta $.

Hint: (b) CD = 5. Draw DE perpendicular to AB, BE = 5, EA = 9.

Q.37. In the following figure, AB = 4 cm and ED = 3 m.

If $\mathrm{sin}\alpha =\frac{3}{5}and\mathrm{cos}\beta =\frac{12}{13}$ find the length of BD.

^{2}+ 4y^{2 }^{}?(a) 2 (b) 1 (c) 0 (d) None of these.

(xxii) If tan $\theta $ + cot $\theta $ = 2, then the value of tan $\theta $ - cot $\theta $ is

(a) 1 (b) 2 (c) - 1 (d) None of these.

(xxiii) If 3 cos $\theta $ - 4 sin $\theta $ = 5, then the value of 3 sin $\theta $ + 4 cos $\theta $ is

(a) 1 (b) 2 (c) 0 (d) None of these.

i) cos angle CBD

ii) cot angle ABD

7. Find AB.

^{o}, AB = 8 cm and BC = 25 cm . Calculate i) BE ii) ACm = a sec A + b tan B

n = a tan A + b sec A

Prove that : m^2 - n^2 = a^2 - b^2

Q12. If $cot\theta =\frac{1}{\sqrt{3}},showthat\left(\frac{1-{\mathrm{cos}}^{2}\theta}{2-{\mathrm{sin}}^{2}\theta}\right)=\frac{3}{5}$

Q13. If $sec\theta =\frac{13}{5},showthat\left(\frac{2\mathrm{sin}\theta -3\mathrm{cos}\theta}{4\mathrm{sin}\theta -9\mathrm{cos}\theta}\right)=3.$

Q14. If $3\mathrm{tan}\theta =4,showthat\left(\frac{3\mathrm{sin}\theta +2\mathrm{cos}\theta}{3\mathrm{sin}\theta -2\mathrm{cos}\theta}\right)=3.$

Q12. Without using trigonometrical tables, evaluate :

(i) $2{\left(\frac{\mathrm{tan}35\xb0}{\mathrm{cot}55\xb0}\right)}^{2}+\left(\frac{\mathrm{cot}55\xb0}{\mathrm{tan}35\xb0}\right)-3\left(\frac{\mathrm{sec}40\xb0}{\mathrm{cosec}50\xb0}\right)$ (ii) $\frac{\mathrm{sin}35\xb0\mathrm{cos}55\xb0+\mathrm{cos}35\xb0\mathrm{sin}55\xb0}{{\mathrm{cosec}}^{2}10\xb0-{\mathrm{tan}}^{2}80\xb0}$

(iii) sin

^{2}34° + sin^{2}56° + 2 tan 18° tan 72° – cot^{2}30°.Q13. Prove the following :

(i) $\frac{\mathrm{cos}\mathrm{\theta}}{\mathrm{sin}\left(90\xb0-\mathrm{\theta}\right)}+\frac{\mathrm{sin}{\displaystyle}{\displaystyle \mathrm{\theta}}}{\mathrm{cos}{\displaystyle}{\displaystyle \left(90\xb0-\mathrm{\theta}\right)}}=2$

(ii) cos $\mathrm{\theta}\mathrm{sin}\left(90\xb0-\mathrm{\theta}\right)+\mathrm{sin}\mathrm{\theta}\mathrm{cos}\left(90\xb0-\mathrm{\theta}\right)=1$

(iii) $\frac{\mathrm{tan}\theta}{\mathrm{tan}\left(90\xb0-\theta \right)}+\frac{{\displaystyle \mathrm{sin}\left(90\xb0-\theta \right)}}{{\displaystyle \mathrm{cos}\theta}}={\mathrm{sec}}^{2}\mathrm{\theta}$.

Q14. Prove the following :

(i) $\frac{\mathrm{cos}\left(90\xb0-\mathrm{A}\right)\mathrm{sin}\left(90\xb0-\mathrm{A}\right)}{\mathrm{tan}\left(90\xb0-\mathrm{A}\right)}=1-{\mathrm{cos}}^{2}\mathrm{A}$

(ii) $\frac{\mathrm{sin}\left(90\xb0-\mathrm{A}\right)}{\mathrm{cosec}\left(90\xb0-\mathrm{A}\right)}+\frac{\mathrm{cos}{\displaystyle}{\displaystyle \left(90\xb0-\mathrm{A}\right)}}{\mathrm{sec}\left(90\xb0-\mathrm{A}\right)}=1$

Q15. Simplify the following :

(i) $\frac{\mathrm{cos}\mathrm{\theta}}{\mathrm{sin}\left(90\xb0-0\right)}+\frac{\mathrm{cos}{\displaystyle}{\displaystyle \left(90\xb0-\mathrm{\theta}\right)}}{\mathrm{sec}\left(90\xb0-\mathrm{\theta}\right)}-3{\mathrm{tan}}^{2}30\xb0$

~~~~ )=1 solve for x1. (Cos 13 + sin 13)/ (cos 13 - sin 13) = tan A

What is A?

2. (sin 20 cos 70 + cos 20 sin 70)/ (sin 23 cosec 23 + cos 23 sec 23)= ?

3. If tan a = 5/6 and tan b = 1/11, then what is the value of a + b ?

~~o~~=6 sin~~o~~, evaluate:~~o~~~~o~~-3 cos~~o~~\12 sin~~o~~+3 cos~~o~~^{2}+ y^{2}= a^{2}+ b^{2}8 (c) In the figure (3) given below, AD is perpendicular to BC, BD=15 cm, sin B= $\frac{4}{5}$ and tan C=1

(i) Calculate the lengths of AD, AB, DC and AC.

Q7. In what time will Rs.2400 amount to Rs. 2646 at 10% p.a. compounded half yearly.

^{2}A=1/(1-cos A)(i) sin A (ii) cos A (iii) tan C

(cosec A/ cosec A - 1) (cosec A/ cosec A + 1) = 2 sec ^2 A

Find

(a)tanx

(b)sinx

Q.4. Prove that

(ii). $4\left({\mathrm{sin}}^{4}30\xb0+{\mathrm{cos}}^{4}60\xb0\right)-3\left({\mathrm{cos}}^{2}45\xb0-{\mathrm{sin}}^{2}90\xb0\right)=2$

$iv.\mathrm{tan}\left(55\xb0-A\right)-cot\left(35\xb0+A\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}v.\mathrm{cos}ec\left(65\xb0+A\right)-sec\left(25\xb0-A\right)$

Q6. If 3$\mathrm{\theta}$ is an acute angle, solve the following equation for $\mathrm{\theta}$ :

(cosec 3$\mathrm{\theta}$ – 2) (cot 2$\mathrm{\theta}$ – 1) = 0.