Prove: (1+tan21)(1+tan28)(1+tan24)(1+tan17)=4
To prove the equation (1+tan(21))(1+tan(28))(1+tan(24))(1+tan(17)) = 4, we will use the fact that tan(θ) + 1 = sec(θ).
1. Begin by substituting sec(θ) - 1 for tan(θ) in the equation. This is because sec(θ) = 1/cos(θ), and tan(θ) = sin(θ)/cos(θ).
(1+tan(21))(1+tan(28))(1+tan(24))(1+tan(17))
= (1+sec(21)−1)(1+sec(28)−1)(1+sec(24)−1)(1+sec(17)−1)
2. Simplify the equation using the identity (A−B)(A+B) = A^2 - B^2.
= [(1+sec(21))^2 − 1^2][(1+sec(28))^2 − 1^2][(1+sec(24))^2 − 1^2][(1+sec(17))^2 − 1^2]
3. Apply the identity sec^2(θ) = 1 + tan^2(θ) to further simplify the equation.
= [sec^2(21) − 1^2][sec^2(28) − 1^2][sec^2(24) − 1^2][sec^2(17) − 1^2]
4. Now we can use the values of sec^2(θ) for each angle:
sec^2(21) = 1.1339
sec^2(28) = 1.2628
sec^2(24) = 1.1547
sec^2(17) = 1.0385
5. Plug in the values and calculate the result:
= [(1.1339 - 1)(1.2628 - 1)(1.1547 - 1)(1.0385 - 1)]
= (0.1339)(0.2628)(0.1547)(0.0385)
≈ 0.001284
6. The result of the left-hand side (LHS) of the equation is approximately 0.001284, which is not equal to 4. Therefore, the equation (1+tan(21))(1+tan(28))(1+tan(24))(1+tan(17)) = 4 is not true.
Hence, the given statement is false.