Question

Show that: 1 + tan2(x) = sec2(x)



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Answer 1

Answer:

$\huge{\bf{\underline{\red{Solution :}}}}$

$\rightarrow$cos²(x)+ sin²(x) = 1

Divide both sides by $\boxed{\orange{cos²(x)}}$ to get:

$\Large\frac{cos²(x)}{cos²(x)} + \frac{sin²(x)}{cos²(x)} = \frac{1}{cos²(x)}$

Which simples to,

$\boxed{\red{1 + tan²(x) = sec²(x)}}$

Answer 2

$\large{\underbrace{\sf{\red{ Correct \: Question :- }}}}$

Show that : 1 + tan²(x) = sec²(x)

$\huge\pink{\mid{{\tt{Solution}}\mid}}$

$\tt \purple{TO \: PROVE :}$

1 + tan²(x) = sec²(x)

$\tt \red{IDENTITY \: USED : }$

cos²(x) + sin²(x) = 1

cot²(x) = 1 + tan²(x)

$\tt \pink{PROOF : }$

As we know,

cos²(x) + sin²(x) = 1

Dividing the equation [cos²(x) + sin²(x) = 1] by cos²(x), we get :-

$\tt \implies\frac{cos²(x)}{cos²(x)} + \frac{sin²(x)}{cos²(x)} = \frac{1}{cos²(x)}$

$\tt \implies \cancel\frac{cos²(x)}{cos²(x)} + \frac{sin²(x)}{cos²(x)} = \frac{1}{cos²(x)}$

$\tt \implies \frac{sin²(x)}{cos²(x)} = \frac{1}{cos²(x)}$

As we know $\tt \frac{sin²(x)}{cos²(x)} \: = {cot}^{2} (x)$ and $\tt\frac{1}{cos²(x)} \: = {sec}^{2} (x)$

so by substituting these value we get :-

$\tt \implies {cot}^{2}(x) = {sec}^{2}(x)$

As we know, cot²(x) = 1 + tan²(x),so substituting 1 + tan²(x) in place of cot²(x) we get :-

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

HENCE PROVED :)