Question

prove

sec^4 A- cot^4 A= 1 + 2tan^2A​

Answer 1

$\texttt{\textsf{\large{\underline{Correct Question}:}}}$

• Prove that – sec⁴(x) - tan⁴(x) = 1 + 2 tan²(x).

$\texttt{\textsf{\large{\underline{Proof}:}}}$

To prove this equality, we have to prove that LHS = RHS.

Taking LHS, we get:

$\sf = {sec}^{4}(x)- {tan}^{4}(x)$

By using identity a² - b² = (a + b) ⋅ (a - b), we get:

$\sf = ( {sec}^{2}x- {tan}^{2}x)( {sec}^{2}x + {tan}^{2}x)$

As we know that:

$\sf \implies{sec}^{2}x- {tan}^{2}x = 1$

We get:

$\sf = {sec}^{2}x + {tan}^{2}x$

Also:

$\sf \implies{sec}^{2}x- {tan}^{2}x = 1$

$\sf \implies{sec}^{2}x = 1 + {tan}^{2}x$

Substituting the value in the expression, we get:

$\sf = (1 + {tan}^{2}x)+ {tan}^{2}x$

$\sf = 1 +2 \: {tan}^{2}x$

We observe that LHS = RHS.

Hence Proved..!!

$\texttt{\textsf{\large{\underline{Learn More}:}}}$

1. Relationship between sides.

• sin(x) = Height/Hypotenuse.
• cos(x) = Base/Hypotenuse.
• tan(x) = Height/Base.
• cot(x) = Base/Height.
• sec(x) = Hypotenuse/Base.
• cosec(x) = Hypotenuse/Height.

2. Square formulae.

• sin²x + cos²x = 1.
• cosec²x - cot²x = 1.
• sec²x - tan²x = 1

3. Reciprocal Relationship.

• sin(x) = 1/cosec(x).
• cos(x) = 1/sec(x).
• tan(x) = 1/cot(x).

4. Cofunction identities.

• sin(90° - x) = cos(x) and vice versa.
• cosec(90° - x) = sec(x) and vice versa.
• tan(90° - x) = cot(x) and vice versa.