Question

Prove That:
$[/tex][tex]\sf\red{ \dfrac{tanθ}{secθ - 1} = \dfrac{tanθ + secθ + 1}{tanθ + secθ - 1}}$

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Answer 1

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

We have to prove that,

$\tt\implies \dfrac{tan(x)}{sec(x)-1}=\dfrac{tan(x)+sec(x)+1}{tan(x)+sec(x)-1}$

We know that,

$\tt\implies sec^{2}(x)-tan^{2}(x)=1$

Therefore,

$\tt\implies sec^{2}(x)-1=tan^{2}(x)$

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

$\tt\implies \{sec(x)+1\}\{sec(x)-1\}=tan(x)\cdot tan(x)$

$\tt\implies \dfrac{sec(x)+1}{tan(x)}=\dfrac{tan(x)}{sec(x)-1} - (i)$

Consider two ratios a/b and c/d, we can say that,

$\tt\implies \dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}$

From (i), we can say that,

$\tt\implies \dfrac{sec(x)+1}{tan(x)}=\dfrac{tan(x)}{sec(x)-1} =\dfrac{tan(x)+sec(x)+1}{tan(x)+sec(x)-1}$

$\tt\implies \dfrac{tan(x)}{sec(x)-1} =\dfrac{tan(x)+sec(x)+1}{tan(x)+sec(x)-1}$

Hence Proved..!!

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

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.