Prove x -n -2 or x - m-2-3 3-8-8- where m - n - I

Sec^2 2x =1 tan 2 x ⇒ 1 +tan^2 2x =1 tan 2x ⇒ tan^2 2x +tan 2x =0 ⇒ tan 2x(tan 2x+1)=0 ⇒ tan 2x =0 or tan 2x = 1 ⇒ 2x =n π or tan 2x = tan π/4=tan ( π π/4 ) ...

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

Byju's Answer

Standard XII

Mathematics

Integration of Trigonometric Functions

Prove x =n π/...

Question

Prove $x = \frac{n π}{2}$ or $x = (\frac{m π}{2} + \frac{3 π}{8})$ , where m, $n \in I$

Open in App

Solution

$s e c^{2} 2 x = 1 - t a n 2 x$

$\Rightarrow 1 + t a n^{2} 2 x = 1 - t a n 2 x$

$\Rightarrow t a n^{2} 2 x + t a n 2 x = 0$

$\Rightarrow t a n 2 x (t a n 2 x + 1) = 0$

$\Rightarrow t a n 2 x = 0$ or $t a n 2 x = - 1$

$\Rightarrow 2 x = n π$ or $t a n 2 x = - t a n \frac{π}{4} = t a n (π - \frac{π}{4}) = t a n \frac{3 π}{4}$

$\Rightarrow x = \frac{n π}{2} o r 2 x = m π + \frac{3 π}{4}$ , where m, $n \in I$

$\Rightarrow x = \frac{n π}{2} o r x = \frac{m π}{2} + \frac{3 π}{8}$ , where m, $n \in I$ .

Suggest Corrections

Similar questions

$x = n π + \frac{3 π}{4}$ or $x = m π + t a n^{- 1} \frac{1}{2}$ , where m, $n \in I$

Prove $x = n π + (- 1)^{n} . \frac{π}{2}$ or $x = (n π + \frac{3 π}{4})$ , where m, $n \in I$

Prove $x = (2 n π \pm \frac{2 π}{3}) o r x = m π + (- 1)^{m} . \frac{7 π}{6}$ , where m, $n \in I$

Q. If

\frac{{cos}^{2} (x + y) - {cos}^{2} (x - y)}{- 4 sin x sin y} = lim x \to 0^{+} sgn(sgn(sgn x))

, where

sgn

is signum function, then which of the following statements is/are correct?

Prove $x = (2 n π + \frac{π}{2})$ or $x = 2 n π$ , where $n \in I$ .

Join BYJU'S Learning Program

Prove x=nπ2 or x=(mπ2+3π8), where m, n∈I

sec22x=1−tan2x ⇒1+tan2 2x=1−tan2x ⇒tan22x+tan2x=0 ⇒tan2x(tan2x+1)=0 ⇒tan2x=0 or tan2x=−1 ⇒2x=nπ or tan2x=−tanπ4=tan(π−π4)=tan3π4 ⇒x=nπ2or2x=mπ+3π4, where m, n∈I ⇒x=nπ2orx=mπ2+3π8, where m, n∈I.

Prove $x = \frac{n π}{2}$ or $x = (\frac{m π}{2} + \frac{3 π}{8})$ , where m, $n \in I$