Question

Hello,

Could somebody guide me to verify:
sinx(tanx+1/tanx)=secx

Answer 1

\(\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\\ \quad \\ \quad \\

sin(x)\left[tan(x)+\cfrac{1}{tan(x)}\right]\implies sin(x)\left[\cfrac{sin(x)}{cos(x)}+\cfrac{1}{\frac{sin(x)}{cos(x)}}\right]
\\ \quad \\
sin(x)\left[\cfrac{sin(x)}{cos(x)}+\cfrac{cos(x)}{sin(x)}\right]\implies ?\)

Answer 2

\[\frac{ \sin ^{2}(x)}{ \cos(x) } +\frac{ \cos(x)\sin(x) }{ \sin(x) } ?\]

Answer 3

well.. is easier if you were to add the fractions inside first

Answer 4

\(\bf sin(x)\left[tan(x)+\cfrac{1}{tan(x)}\right]\implies sin(x)\left[\cfrac{sin(x)}{cos(x)}+\cfrac{1}{\frac{sin(x)}{cos(x)}}\right]
\\ \quad \\
sin(x)\left[\cfrac{sin(x)}{cos(x)}+\cfrac{cos(x)}{sin(x)}\right]\implies sin(x)\left[\cfrac{sin^2(x)+cos^2(x)}{cos(x)sin(x)}\right]
\\ \quad \\
{\color{brown}{ recall\implies sin^2(\theta)+cos^2(\theta)=1}}\qquad thus
\\ \quad \\
sin(x)\left[\cfrac{sin^2(x)+cos^2(x)}{cos(x)sin(x)}\right]\implies \cancel{ sin(x) }\left[\cfrac{1}{cos(x)\cancel{ sin(x) }}\right]\)

Answer 5

thank you so much, you were really helpful

Answer 6

yw