Question

one upon sec theta minus tan theta equal to sec theta plus tan theta

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to prove that:

$\rm \longrightarrow \dfrac{1}{ \sec(x) - \tan(x) } = \sec(x) + \tan(x)$

We know that:

$\rm \longrightarrow \sec^{2}(x) - \tan^{2}(x) = 1$

Using identity a² - b² = (a + b)(a - b), we get:

$\rm \longrightarrow [ \sec(x) + \tan(x) ][ \sec(x) - \tan(x) ]= 1$

$\rm \longrightarrow \sec(x) + \tan(x) = \dfrac{1}{ \sec(x) - \tan(x)}$

Hence Proved..!!

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

1. Relationship between sides and T-Ratios.

• 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.

• sin²(x) + cos²(x) = 1
• cosec²(x) - cot²(x) = 1
• sec²(x) - tan²(x) = 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)
• cos(90° - x) = sin(x)
• cosec(90° - x) = sec(x)
• sec(90° - x) = cosec(x)
• tan(90° - x) = cot(x)
• cot(90° - x) = tan(x)

5. Even odd identities.

• sin(-x) = -sin(x)
• cos(-x) = cos(x)
• tan(-x) = -tan(x)