Show that 1-tan2x-2xwhere x- 2n-1-2- n - I

Solving LHS, we get 1+tan^2x 1+sin^2xcos^2x cos^2x+sin^2xcos^2x 1cos^2x ^2x #160; = RHS,Hence proved ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Trigonometric Identities

Show that 1...

Question

Show that $1 + {tan}^{2} x = {sec}^{2} x$
where $x \neq (2 n + 1) \frac{π}{2}, n ϵ I$

Open in App

Solution

Suggest Corrections

Similar questions

2 s i n^{2} ((\frac{π}{2}) c o s^{2} x) = 1 - c o s (π s i n 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

c o s 2 x

2 {sin}^{2} ((\frac{π}{2}) {cos}^{2} x) = 1 - cos (π {sin}^{2} 2 x), x \neq (2 n + 1) \frac{π}{2}, n ϵ I

cos 2 x

1 + {cot}^{2} x = {csc}^{2} x

x \neq n π, n ϵ I

L e t f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} (1 + | c o s x |)^{\frac{a b}{| c o s x |}}, & n π < x < (2 n + 1) \frac{π}{2} e^{a} . e^{b}, & x = (2 n + 1) \frac{π}{2} e^{\frac{c o t 2 x}{c o t 8 x}}, & (2 n + 1) \frac{π}{2} < x < (n + 1) π \end{matrix} I f f (x) i s c o n t i n u o u s i n ((n π), (n + 1) π, t h e n)

sec x - 1 = (\sqrt{2} - 1) tan x, x \neq (2 n - 1) \frac{π}{2}, n \in Z

Join BYJU'S Learning Program