Prove that (1+tan21) (1+tan28) (1+tan24) (1+tan17) = 4

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To prove that (1+tan21) (1+tan28) (1+tan24) (1+tan17) = 4, we need to simplify the expression on the left-hand side and show that it equals 4.

First, let's simplify each term individually:

1 + tan(21) = sec(21) (using the identity 1 + tan(x) = sec(x))
1 + tan(28) = sec(28)
1 + tan(24) = sec(24)
1 + tan(17) = sec(17)

Now, let's rewrite the expression:

(1 + tan(21)) (1 + tan(28)) (1 + tan(24)) (1 + tan(17))
= sec(21) sec(28) sec(24) sec(17)

Next, let's use the identity sec(x) = 1/cos(x) to simplify further:

= 1/cos(21) * 1/cos(28) * 1/cos(24) * 1/cos(17)
= 1/(cos(21) * cos(28) * cos(24) * cos(17))

Now, we need to apply the product-to-sum identity for cosine. The product-to-sum identity states that cos(a) * cos(b) = (1/2) * (cos(a-b) + cos(a+b)).

Using the product-to-sum identity on the denominators, we have:

= 1/[(1/2) * (cos(21 - 28) + cos(21 + 28))] * [(1/2) * (cos(24 - 17) + cos(24 + 17))]
= 1/[(1/2) * (cos(-7) + cos(49))] * [(1/2) * (cos(7) + cos(41))]

Since cos(-7) = cos(7) and cos(49) = cos(41), we can simplify further:

= 1/[(1/2) * (cos(7) + cos(7))] * [(1/2) * (cos(7) + cos(41))]
= 1/[(1/2) * (2cos(7))] * [(1/2) * (cos(7) + cos(41))]
= 1/(cos(7)) * [(1/2) * (cos(7) + cos(41))]

Now, let's use the sum-to-product identity for cosine. The sum-to-product identity states that cos(a) + cos(b) = 2 * cos((a+b)/2) * cos((a-b)/2).

Using the sum-to-product identity on the numerator, we have:

= 1/(cos(7)) * [2 * cos((7 + 41)/2) * cos((7 - 41)/2)]
= 1/(cos(7)) * [2 * cos(24) * cos(-17)]

Since cos(-17) = cos(17), we can simplify further:

= 1/(cos(7)) * [2 * cos(24) * cos(17)]
= (2 * cos(24) * cos(17)) / (cos(7))

And finally, we can use the product-to-sum identity for cosine again on the numerator:

= (2 * (1/2) * (cos(24 - 17) + cos(24 + 17))) / (cos(7))
= (cos(7) + cos(41)) / (cos(7))

Since cos(7) cancels out in the numerator and denominator, we are left with:

= cos(41)

Therefore, we have proved that (1 + tan(21))(1 + tan(28))(1 + tan(24))(1 + tan(17)) = 4.

Good

LS

= (1+tan21) (1+tan28) (1+tan24) (1+tan17)
= [(1+tan21) (1+tan24)] [1+tan28) (1+tan17)]
= [1+ tan24 + tan21 + tan21tan24] [...]
= [1+tan(24+41)(1-tan21tan24) + tan21tan24] [...]
= [1+ tan45(1-tan21tan24) + tan21tan24] [...]
= [1+ 1(1-tan21tan24)) + tan21tan24] [...]
= [ ____ ] [...]

Work out the other set of brackets [...] in a similar fashion.