Prove the following identities: (i) (sin x + cos x)(tan x + cot x) = sec x + cosec x

(ii) 1/(sec x - tan x) - tan x = sec x​

Step-by-step explanation:

Left-hand-side:

\displaystyle \dfrac{1}{sec(x) - tan(x)}

sec(x)−tan(x)

1

\displaystyle = \ \dfrac{1}{sec(x) - tan(x)} * \dfrac{sec(x) + tan(x)}{sec(x) + tan(x)}=

sec(x)−tan(x)

1

∗

sec(x)+tan(x)

sec(x)+tan(x)

\displaystyle = \ \dfrac{sec(x) + tan(x)}{sec^2(x) - tan^2(x)}=

sec

2

(x)−tan

2

(x)

sec(x)+tan(x)

Now use

\displaystyle 1 + tan^2(x) \ = \ sec^2(x)1+tan

2

(x) = sec

2

(x)

and you should be done

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i ran through it by the other answer is wrong