Question

(tanA×sinA)+cosA=secA​

Answer 1

Correct Question:

• Prove that: tan(x) sin(x) + cos(x) = sec(x)

Proof:

Taking LHS,

As we know that tan(x) = sin(x)/cos(x), So,

$\rm \tan(x) \sin(x) + \cos(x)$

$\rm = \dfrac{ \sin(x) }{ \cos(x) } \times \sin(x) + \cos(x)$

$\rm = \dfrac{ { \sin}^{2}(x)}{ \cos(x) } + \cos(x)$

$\rm = \dfrac{ { \sin}^{2}(x) + { \cos}^{2}(x)}{ \cos(x) }$

We know that,

$\rm \mapsto { \sin}^{2}(x) + \cos^{2} (x) = 1$

Therefore,

$\rm \dfrac{ { \sin}^{2}(x) + { \cos}^{2}(x)}{ \cos(x) }$

$\rm = \dfrac{1}{ \cos(x) }$

As sec(x) is the opposite of cos(x), So,

$\rm \dfrac{1}{ \cos(x) }$

$\rm = \sec(x)$

= RHS (Hence Proved)

Note:

• tan(x) = sin(x)/cos(x)
• sec(x) = 1/cos(x)
• sin²(x) + cos²(x) = 1