Question

$\frac{\sec 8\text{A}-1}{\sec 4\text{A}-1}=$

• A. $\frac{\tan 2\text{A}}{\tan 8\text{A}}$
• B. $\frac{\tan 8\text{A}}{\tan 2\text{A}}$
• C. $\frac{\text{cot8A}}{\cot 2\text{A}}$
• D. $\frac{\text{tan6A}}{\tan 2\text{A}}$

Answer 1

Answer: (B)

$\frac{\tan 8\text{A}}{\tan 2\text{A}}$

Solution :-

$\frac{sec8A-1}{sec4A-1}=\frac{1/cos8A-1}{1/cos4A-1}$

$\Rightarrow \frac{\frac{1-cos8A}{cos8A}}{\frac{1-cos4A}{cos4A}}=\frac{cos4A(1-cos8A)}{cos8A(1-cos4A)}$

$\Rightarrow \frac{cos4A(1-(1-2si{n}^{2}4A\left)\right)}{cos8A(1-(1-2si{n}^{2}2A\left)\right)}=\frac{cos4A.si{n}^{2}2A}{cos8A.si{n}^{2}2A}$

$\Rightarrow \frac{\left(2cos4Asin4A\right)sin4A}{\left(2cos8Asi{n}^{2}2A\right)}=\frac{sin8Asin4A}{2cos8A.si{n}^{2}2A}$

$\Rightarrow tan8A\left(\frac{sin4A}{2si{n}^{2}A}\right)=tan8A.\left(\frac{2sin2A.cos2A}{2si{n}^{2}2A}\right)$

$\Rightarrow \frac{tan8A}{tan2A}$