f(x)=ln(tan(x)+sec(x))-ln(1-sin(x)) −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 −3 -2.5 −2 -1.5 −1 -0.5 0 0.5 1 1.5 2 2.5 3 x y
Exerciseln ( tan ( x ) + sec ( x ) ) − ln ( 1 − sin ( x ) ) \ln\left(\tan\left(x\right)+\sec\left(x\right)\right)-\ln\left(1-\sin\left(x\right)\right) ln ( tan ( x ) + sec ( x ) ) − ln ( 1 − sin ( x ) )
Step-by-step Solution
1
The difference of two logarithms of equal base b b b log b ( x ) − log b ( y ) = log b ( x y ) \log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\right) log b ( x ) − log b ( y ) = log b ( y x )
ln ( tan ( x ) + sec ( x ) 1 − sin ( x ) ) \ln\left(\frac{\tan\left(x\right)+\sec\left(x\right)}{1-\sin\left(x\right)}\right) ln ( 1 − sin ( x ) tan ( x ) + sec ( x ) )
Final answer to the exercise ln ( tan ( x ) + sec ( x ) 1 − sin ( x ) ) \ln\left(\frac{\tan\left(x\right)+\sec\left(x\right)}{1-\sin\left(x\right)}\right) ln ( 1 − sin ( x ) tan ( x ) + sec ( x ) )