Question

The expression secX + tanX is equal to -

- tan [π/4 + x/2]

- tan [π/4 - x/2]

- tan x/2

- tan x/4

Which one is correct?

Answer 1

Answer:

3rd one is correct .. hope this helps

Answer 2

$\large\underline{\sf{Solution-}}$

Given Trigonometric function is

$\rm :\longmapsto\:secx + tanx$

$\rm \:  =  \: \dfrac{1}{cosx} + \dfrac{sinx}{cosx}$

$\rm \:  =  \: \dfrac{1 + sinx}{cosx}$

can be rewritten as

$\rm \:  =  \: \dfrac{1 + cos\bigg[\dfrac{\pi}{2} - x\bigg]}{sin\bigg[\dfrac{\pi}{2} - x \bigg]}$

Let assume that,

$\red{ \boxed{ \sf{ \:\dfrac{\pi}{2} - 2x = y}}}$

$\rm \:  =  \: \dfrac{1 + cosy}{siny}$

We know,

$\red{ \boxed{ \sf{ \:1 + cos2x = {2cos}^{2}x}}}$

and

$\red{ \boxed{ \sf{ \:sin2x = 2sinx \: cosx}}}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{2 {cos}^{2}\bigg[\dfrac{y}{2} \bigg] }{2sin\bigg[\dfrac{y}{2} \bigg]cos\bigg[\dfrac{y}{2} \bigg]}$

$\rm \:  =  \: cot\dfrac{y}{2}$

On Substituting the value of y, we get

$\rm \:  =  \: cot\bigg[\dfrac{\pi}{4} - \dfrac{x}{2} \bigg]$

We know,

$\red{ \boxed{ \sf{ \:cot\bigg[\dfrac{\pi}{2} - x \bigg] = tanx}}}$

So, using this we get,

$\rm \:  =  \: tan\bigg[\dfrac{\pi}{2} - \dfrac{\pi}{4} + \dfrac{x}{2} \bigg]$

$\rm \:  =  \: tan\bigg[\dfrac{\pi}{4} + \dfrac{x}{2} \bigg]$

Hence,

$\red{ \boxed{ \sf{ \:secx + tanx=  \: tan\bigg[\dfrac{\pi}{4} + \dfrac{x}{2} \bigg]}}}$

• So, Option (1) is correct

Additional Information :-

$\red{ \boxed{ \sf{ \:sin2x = \frac{2tanx}{1 + {tan}^{2}x }}}}$

$\red{ \boxed{ \sf{ \:tan2x = \frac{2tanx}{1 - {tan}^{2}x }}}}$

$\red{ \boxed{ \sf{ \:cos2x = \frac{1 - {tan}^{2} x}{1 + {tan}^{2}x }}}}$

$\red{ \boxed{ \sf{ \:cos2x = {cos}^{2}x - {sin}^{2}x = 1 - {2sin}^{2}x = {2cos}^{2}x - 1}}}$

$\red{ \boxed{ \sf{ \:sin3x = 3sinx - {4sin}^{3}x}}}$

$\red{ \boxed{ \sf{ \:cos3x = {4cos}^{3}x - 3cosx}}}$