Question

1) secΘ = 5/7, Find 1 + tanΘ/1 - tanΘ
2) Prove that, 2 cos²Θ - 1/1- 2sin²Θ = 1
3) sinA = 3/5, cosB = 12/13, Find tanA - tanB/1 + tanA - tanB
4) If sin(α - β) = 1, cos (2α + β) = 1/2, Find tanα​

Answer 1

Question 1) :- secΘ = 5/4 Find 1 + tanΘ/1 - tanΘ .

we know that,

• sec²A - tan²A = 1 .

So,

→ sec²Θ - tan²Θ = 1

→ (5/4)² - tan²Θ = 1

→ (25/16) - 1 = tan²Θ

→ tan²Θ = (25 - 16)/16

→ tan²Θ = 9/16

→ tanΘ = 3/4 .

Putting value now,

→ (1 + 3/4) / (1 - 3/4)

→ (7/4) / (1/4)

→ 7 (Ans.)

________________

Question :- 2) 2 cos²Θ - 1/1- 2sin²Θ = 1 .

Solution :-

we know that,

• sin²A + cos²A = 1 .
• sin²A = (1 - cos²A)

Putting value of sin²Θ in denominator , we get,

→ (2cos²Θ - 1) / {1 - 2((1 - cos²Θ)} = 1 .

→ (2cos²Θ - 1) / ( 1 - 2 + 2cos²Θ) = 1

→ (2cos²Θ - 1) / (2cos²Θ - 1) = 1

→ 1 = 1 . (Proved.)

________________

Question 3) :- sinA = 3/5, cosB = 12/13, Find tanA - tanB/1 + tanA - tanB .

Solution :-

we know that,

• sinA = Perpendicular / Hypotenuse
• cosA = Base / Hypotenuse .
• tanA = Perpendicular/Base .

So,

→ sinA = 3/5 = P/H

• P = 3
• H = 5

then,

→ B = √[H² - P²] = √[5² - 3²] = √(25 - 9) = √16 = 4.

Similarly,

→ cosB = 12/13 = B/H

• B = 12
• H = 13

then,

→ P = √[H² - B²] = √[13² - 12²] = √(169 - 144) = √25 = 5.

therefore,

• tanA = P/B = 3/4 .
• tanB = P/B = 5/12 .

Hence,

→ (tanA - tanB)/(1 + tanA - tanB)

→ (3/4 - 5/12) / (1 + 3/4 - 5/12)

→{(9 - 5) /12} / {(12 + 9 - 5)/12}

→ (4/12) / (16/12)

→ (4/12) * (12/16)

→ (1/4) (Ans.)

_____________________

Question 4) :- If sin(α - β) = 1, cos (2α + β) = 1/2, Find tanα ?

we know that,

• sin90° = 1 .
• cos60° = 1/2 .

So,

→ sin(α - β) = 1

→ sin(α - β) = sin90°

→ (α - β) = 90° --------- Eqn.(1) .

and,

→ cos (2α + β) = 1/2

→ cos (2α + β) = cos60°

→ (2α + β) = 60° --------- Eqn.(2)

adding Eqn.(1) and Eqn.(2) ,

→ (α - β) + (2α + β) = 60° + 90°

→ 3α = 150°

→ α = 50° .

therefore,

→ tanα = tan50°

_________________

Answer 2

Answer:

please follow me my dear friend please ❤️❤️