learn this (Consider the expression $ e^x-ln(cos^2(x))+ln(1+tan^2(x))$ and then simplify it into a single term

Solution We are given the expression:

e x − ln ⁡ ( cos ⁡ 2 ( x ) ) + ln ⁡ ( 1 + tan ⁡ 2 ( x ) ) . e x −ln(cos 2 (x))+ln(1+tan 2 (x)). Let's simplify this step by step.

Step 1: Simplifying ln ⁡ ( cos ⁡ 2 ( x ) ) ln(cos 2 (x)) We use the logarithmic property ln ⁡ ( a b )

b ln ⁡ ( a ) ln(a b )=bln(a), where a

cos ⁡ ( x ) a=cos(x) and b

2 b=2. Thus,

ln ⁡ ( cos ⁡ 2 ( x ) )

2 ln ⁡ ( cos ⁡ ( x ) ) . ln(cos 2 (x))=2ln(cos(x)). So the expression becomes:

e x − 2 ln ⁡ ( cos ⁡ ( x ) ) + ln ⁡ ( 1 + tan ⁡ 2 ( x ) ) . e x −2ln(cos(x))+ln(1+tan 2 (x)). Step 2: Simplifying ln ⁡ ( 1 + tan ⁡ 2 ( x ) ) ln(1+tan 2 (x)) Recall the trigonometric identity 1 + tan ⁡ 2 ( x )

sec ⁡ 2 ( x ) 1+tan 2 (x)=sec 2 (x). Using this identity, we have:

ln ⁡ ( 1 + tan ⁡ 2 ( x ) )

ln ⁡ ( sec ⁡ 2 ( x ) )

2 ln ⁡ ( sec ⁡ ( x ) ) . ln(1+tan 2 (x))=ln(sec 2 (x))=2ln(sec(x)). Now the expression becomes:

e x − 2 ln ⁡ ( cos ⁡ ( x ) ) + 2 ln ⁡ ( sec ⁡ ( x ) ) . e x −2ln(cos(x))+2ln(sec(x)). Step 3: Using the identity sec ⁡ ( x )

1 cos ⁡ ( x ) sec(x)= cos(x) 1 ​

Next, we use the fact that sec ⁡ ( x )

1 cos ⁡ ( x ) sec(x)= cos(x) 1 ​ . Therefore, we can rewrite ln ⁡ ( sec ⁡ ( x ) ) ln(sec(x)) as:

ln ⁡ ( sec ⁡ ( x ) )

ln ⁡ ( 1 cos ⁡ ( x ) )

− ln ⁡ ( cos ⁡ ( x ) ) . ln(sec(x))=ln( cos(x) 1 ​ )=−ln(cos(x)). Substituting this into the expression, we get:

e x − 2 ln ⁡ ( cos ⁡ ( x ) ) + 2 ( − ln ⁡ ( cos ⁡ ( x ) ) )

e x − 2 ln ⁡ ( cos ⁡ ( x ) ) − 2 ln ⁡ ( cos ⁡ ( x ) ) . e x −2ln(cos(x))+2(−ln(cos(x)))=e x −2ln(cos(x))−2ln(cos(x)). This simplifies further to:

e x − 4 ln ⁡ ( cos ⁡ ( x ) ) . e x −4ln(cos(x)). Final Simplified Expression: Thus, the simplified expression is:

e x − 4 ln ⁡ ( cos ⁡ ( x ) ) . e x −4ln(cos(x)).)