1 . সমাধান করুনঃ tan2θ-(1+√3)tanθ+√3 = 0

• A. θ = 15°, 30°
• B. θ = 45°, 60°
• C. θ = 36°, 55°
• D. θ = 75°, 90°

Answer: Option B

Explanation:

2 . A=30°

• A. $2$
• B. $\frac{2}{\sqrt{3}}$
• C. $4$
• D. $2\sqrt{3}$

Answer: Option D

Explanation:

3 . 11+tan2A+11+cos2A

• A. $\frac{1}{\sin A}$
• B. $1$
• C. $\frac{1}{\cos A}$
• D. $\frac{1}{\tan A}$

Answer: Option C

Explanation:

4 . 1+tan2θ=4

• A. $30°$
• B. $45°$
• C. $60°$
• D. $0°$

Answer: Option C

Explanation:

5 . tan2

• A. থিটা=$15°,30°$
• B. থিটা=$45°,60°$
• C. থিটা=$36°,55°$
• D. থিটা=$75°,90°$

Answer: Option B

Explanation:

6 . tan2θtanθ=1

• A. 0°
• B. 30°
• C. 45°
• D. 60°

Answer: Option B

Explanation:

7 . limx→0ex-1tan2xx3=

• A. -1
• B. 1
• C. 4
• D. 3

Answer: Option B

Explanation:

8 . θ=cos-145

• A. $\frac{16}{25}$
• B. $\frac{9}{25}$
• C. $\frac{8}{25}$
• D. $\frac{7}{25}$

Answer: Option D

Explanation:

9 . 5tan2θ-sec2θ=11n<θ<2pi

• A. $\frac{7\mathrm{\pi }}{3}$
• B. $\frac{2\mathrm{\pi }}{3}$
• C. $\frac{4\mathrm{\pi }}{3}$
• D. $\frac{\mathrm{\pi }}{3}$

Answer: Option D

Explanation:

10 . tan2θtanθ=1

• A. 0°
• B. 30°
• C. 45°
• D. 60°

Answer: Option B

Explanation:

11 . 1-tan2A1+tan2A

• A. 1
• B. $\frac{1}{2}$
• C. 0
• D. 2

Answer: Option C

Explanation:

12 . θ=60°

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $1$

Answer: Option D

Explanation:

13 . 2tanθ1+tan2θ=?

• A. $\tan z^2\theta$
• B. $2 {\cos }^2\frac{\theta }{2}$
• C. $2 \sin \theta \cos \theta$
• D. $\cos 2\theta$

Answer: Option C

Explanation:

14 . tan2θ+secθ=-1;0<θ<2π

• A. $\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{3\mathrm{\pi }}{2}$

Answer: Option A

Explanation:

15 . cosθ=1213

• A. $\pm \frac{120}{119}$
• B. $\pm \frac{2}{12}$
• C. $\pm \frac{144}{169}$
• D. $\pm \frac{25}{144}$

Answer: Option A

Explanation: