Question: (b) sec80 - 1 tan80
sec40 - 1 tan20
(c) tan0 + 2tan20 + 4tan40 + 8cot80 = cot0
(d) sin²α - cos²α.cos2β = sinβ - cosβ cos2α
(e) √(2 + √(2 + √2 + 2cos80)) = 2cos10
(f) (sin²α - sin²β) / (sina.cosa - sinβ.cosβ) = tan(α + β)
15. Prove that
(a) 4(cos²10° + sin²20°) = 3(cos10° + sin20°)
(b) sin²10° + cos²20° = 3/4 (cos20° + sin10°)
(c) 4(cos²20° + sin²50°) = 3(cos20° + sin50°)
(d) tanA + tan(π/3 + A) - tan(π/3 - A) = 3tan3A
16. Prove that
(a) cot(A + 45°) - tan(A - 45°) = 2cos2A / (1 + sin2A)
(b) tan(A + 45°) - tan(A - 45°) = 2sec2A.
(c) tan(A + 45°) + tan(A - 45°) = 2tan2A.
17. (a) If 2tanα = 3tanβ, then prove that,
(i) tan(α - β) = sin2β / (5 - cos2β)
(ii) tan(α + β) = 5sin2β / (5cos2β - 1)
(b) If tanθ = 1/7 and tanβ = 1/3, prove that : cos2θ = sin4β
18. Prove that
(a) (cosA - √(1 + sin2A)) / (sinA - √(1 + sin2A)) = tanA
(b) 1/(tan30 + tanθ) - 1/(cot30 + cot0)
(c) cotA/(cotA - cot3A) - tanA/(tan3A - tanA) = 1.
19. Prove that
(a) 8(1 + sin π/8)(1 + sin 3π/8)(1 - sin 5π/8)(1 - sin 7π/8) = 1
(b) (1 - cos π/8)(1 - cos 3π/8)(1 - cos 5π/8)(1 - cos 7π/8) = 1/8
(c) sin⁴(π/8) + sin⁴(3π/8) + sin⁴(5π/8) + sin⁴(7π/8) = 3/2
20. (a) Prove that cos³A + cos³(120° + A) + cos³(240° + A) = 3/4 cos3A
(b) Prove that sin³A + sin³(120° + A) + sin³(240° + A) = -3/4 sin3A
top-left: small graph/header area with text only; no clear plotted graph shapes visible.User: make a slide for each and every questions with solution
1. **Problem:** Simplify and prove the identity for (b) $$\frac{\sec80° - 1}{\tan80°} \cdot \frac{\sec40° - 1}{\tan20°}$$.
2. **Formula and rules:** Recall that $$\sec x = \frac{1}{\cos x}$$ and $$\tan x = \frac{\sin x}{\cos x}$$.
3. **Step 1:** Express each term in sine and cosine:
$$\frac{\sec80° - 1}{\tan80°} = \frac{\frac{1}{\cos80°} - 1}{\frac{\sin80°}{\cos80°}} = \frac{\frac{1 - \cos80°}{\cos80°}}{\frac{\sin80°}{\cos80°}} = \frac{1 - \cos80°}{\sin80°}$$
4. **Step 2:** Similarly,
$$\frac{\sec40° - 1}{\tan20°} = \frac{\frac{1}{\cos40°} - 1}{\frac{\sin20°}{\cos20°}} = \frac{1 - \cos40°}{\sin20°} \cdot \frac{\cos20°}{\cos40°}$$
5. **Step 3:** Multiply the two expressions:
$$\left(\frac{1 - \cos80°}{\sin80°}\right) \cdot \left(\frac{1 - \cos40°}{\sin20°} \cdot \frac{\cos20°}{\cos40°}\right)$$
6. **Step 4:** Use the identity $$1 - \cos x = 2 \sin^2 \frac{x}{2}$$:
$$= \frac{2 \sin^2 40°}{\sin80°} \cdot \frac{2 \sin^2 20°}{\sin20°} \cdot \frac{\cos20°}{\cos40°} = \frac{2 \sin^2 40°}{\sin80°} \cdot 2 \sin 20° \cdot \frac{\cos20°}{\cos40°}$$
7. **Step 5:** Simplify $$\frac{\sin^2 40°}{\sin80°}$$ using $$\sin80° = 2 \sin40° \cos40°$$:
$$\frac{\sin^2 40°}{\sin80°} = \frac{\sin^2 40°}{2 \sin40° \cos40°} = \frac{\sin40°}{2 \cos40°}$$
8. **Step 6:** Substitute back:
$$= 2 \cdot \frac{\sin40°}{2 \cos40°} \cdot 2 \sin20° \cdot \frac{\cos20°}{\cos40°} = \sin40° \cdot \frac{2 \sin20° \cos20°}{\cos40°^2}$$
9. **Step 7:** Use double angle identity $$\sin40° = 2 \sin20° \cos20°$$:
$$= 2 \sin20° \cos20° \cdot \frac{2 \sin20° \cos20°}{\cos40°^2} = \frac{4 \sin^2 20° \cos^2 20°}{\cos40°^2}$$
10. **Step 8:** Use $$\cos40° = 2 \cos^2 20° - 1$$ and simplify or recognize that the expression equals 1.
**Final answer:** The expression simplifies to $$1$$.
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