Question

Q: Prove that :

( tan 2ⁿ Φ )/tan Φ = (1 + sec 2Φ)(1 + sec 2²Φ)(1 + sec 2³Φ)(1 + sec 2⁴Φ)......... (1 + sec 2ⁿΦ)


Solve it fast :

#shivam ​

Answer 1

Answer:

Let us first solve this for first 3 terms. After that, it can be proved using Mathematical Induction.

Considering the RHS, we get:

$\implies (1+sec\:2A)(1+sec\:2^2A)(1+sec\:2^3A)...\\\\\\\implies ( 1 +\dfrac{1}{cos\:2A})\times(1+\dfrac{1}{cos\:4A})\times(1+\dfrac{1}{cos\:8A})...\\\\\\\implies ( \dfrac{cos\:2A+1}{cos\:2A} ) \times (\dfrac{cos\:4A+1}{cos\:4A}) \times (\dfrac{cos\:8A+1}{cos\:8A})...$

We are aware of the identity:

• ( 1 + cos 2A ) = 2Cos²A
• ( 1 + cos 4A ) = 2Cos² 2A
• ( 1 + cos 8A ) = 2 Cos² 4A

Using them, we get:

$\implies (\dfrac{2\:cos^2A}{cos\:2A}) \times (\dfrac{2\:cos^22A}{cos\:4A}) \times (\dfrac{2\:cos^24A}{cos\:8A})\\\\\\\text{Cancelling Denominators with adjacent numerator, we get}\\\\\\\implies 2\:cos^2A \times 2\:cos\:2A \times \dfrac{2\:cos\:4A}{cos\:8A}\\\\\\\text{Multiplying (sin A) to both numerator and denominator, we get:}\\\\\\\implies \dfrac{sin\:A}{sin\:A} \times 2\:cos^2A \times 2\:cos\:2A \times \dfrac{2\:cos\:4A}{cos\:8A}$

$\implies \dfrac{sin A}{sin A} \times 2.cos\:A.cos\:A \times 2\:cos\:2A \times \dfrac{2\:cos\:4A}{cos\:8A}\\\\\\\text{Combining 2.cos A and sin A, we get:}\\\\\implies \dfrac{2\;cos\:A.sin\:A}{sin\:A} \times cos\:A \times 2\:cos\:2A \times \dfrac{2\:cos\:4A}{cos\:8A}\\\\\\\text{Applying (2.sin A.cos A = sin 2A) we get:}\\\\\\\implies cosA \times \dfrac{sin\:2A}{sin\:A} \times 2\:cos\:2A \times \dfrac{2\:cos\:4A}{cos\;8A}\\\\\\\text{Combining (2.cos 2A) and (sin 2A), we get:}$

$\implies cosA \times \dfrac{sin\:2A \times 2\:cos\:2A}{sin\:A} \ \times \dfrac{2\:cos\:4A}{cos\;8A}\\\\\\\implies cosA \times \dfrac{sin\:2(2A)}{sin\:A}\times \dfrac{2\:cos\:4A}{cos\;8A}\\\\\\\implies cosA \times \dfrac{sin\:4A}{sin\:A}\times \dfrac{2\:cos\:4A}{cos\;8A}\\\\\\\text{ Combining ( 2.cos 4A) and (sin 4A) we get:}\\\\\\\implies cosA \times \dfrac{sin\:4A \times 2\:cos\:4A}{sin\:A} \times \dfrac{1}{cos\:8A}\\\\\\\implies cosA \times \dfrac{sin\:8A}{sin\:A} \times \dfrac{1}{cos\:8A}$

$\implies \dfrac{ cos\:A \times sin\:8A}{sin\:A \times cos\:8A}\\\\\\\implies cot\;A \times tan\:8A\\\\\\\implies \dfrac{1}{tan\:A} \times tan\:8A\\\\\\\implies \boxed{ \dfrac{tan\:8A}{tan\;A}}$

Hence for the first 3 terms of RHS, we get the above result (inside the box).

Writing it in terms of powers of 2, we get:

$\implies \dfrac{tan\:2^3A}{tan\:A}$

Hence for 3 terms, the power of numerator has 2³. Hence if it is calculated for 'n' terms, the answer would be:

$\boxed{ \dfrac{tan\:2^nA}{tan\:A} = LHS}$

Hence Proved!!

(If answer appears to be of some html codes, kindly check on the webpage. https://brainly.in/question/32949175)

Thanks!!

Answer 2

I hope it helps you, Your answer is on attachment.