Question

1. Evaluate $\frac{2cot45°}{1+tan45°}$

2. Evaluate $\frac{cos45°}{sec30° + cosec60°}$

3. Evaluate sin90°+cos0°+√2cos45°

4. compute $\frac{sin37°}{cos53°}$ - 2tan45° + $\frac{cot47°}{tan43°}$

5. If sinθ = $\frac{5}{3}$. Evaluate $\frac{1+sinθ}{1-sinθ}$

6. if sin(A-B) = $\frac{1}{2}$ , cos(A+B) = $\frac{1}{2}$, then show that 0° < A+B ≤ 90°
7. If sinA = cosB. show that A+B = 90°

8. If secθ+tanθ = p then show :
a) sinA : $\frac{p²+1}{p²-1}$
b) If A,B,C are three angles of a ΔABC, prove that tan $\frac{B+C}{2}$ = cot $\frac{C}{r}$

9. a) given tanθ = $\frac{15}{8}$, evaluate $\frac{17sinθ + 15cotθ}{16secθ + 8tanθ}$
b) A chord of a circle of radius 6cm is making 60° at the center. Find length of the chord.
c) Solve $\frac{cosθ}{1-sinθ}$ + $\frac{cosθ}{1+sinθ}$ = 4
d) If radius of a circle is 14cm. The angle subtracted by an arc of the circle at the center. Find the length of arc.

10. 75° is an :
a. Acute angle
b. Obtuse angle
c. right angle
d. straight angle

11. If tanθ : cotθ ; θ = ______
a. 90°
b. 45°
c. 60°
d. 30°

12. sin²65° + sin²65° = _________
a. 2
b. 1
c. -1
d. -2

Answer 1

9(d) length of arc = (θ/360°)×2πr
 substitute the value of θ and r

10. 75°is an acute angle as 0<75°<90°

11. It is tanθ = cotθ or tanθ:cotθ = 1:1
Both are same thing,

tanθ = 1/tanθ
⇒tan²θ = 1
⇒θ = 45°

12.It should be sin²65°+cos²65°
= 1
∵sin²θ +cos²θ = 1 for any value of θ

Rest answers are attached.
Doubts are most welcomed.