Question

sin²x=1-tan2x genra formulat​

Answer 1

Step-by-step explanation:

$\tt\implies sin^{2}x = 1 - tan2x$

we know that ,

$\tt\implies tan2x = \frac{2tanx}{1 - {tan}^{2}x }$

Now ,

$\tt\circ\: sin^{2}x = 1 - tan2x \\ \\ \tt \implies 1 - {sin}^{2} x \: =tan2x \\ \\ \bold{we \: know \: that} \: \\ \tt \implies 1 - {sin}^{2} x \: = \: {cos}^{2} x \\ \\ \bold{now} \\ \\ \tt\implies {cos}^{2} x \: = tan2x \\ \\ \tt\implies {cos}^{2} x = \frac{2tan\:x}{1 - {tan}^{2}x } \\\\ \tt\implies cos^{2}x ( 1 - tan^{2}x ) = 2tan\:x\\\\ \tt\implies cos^{2}x - cos^{2}x tan^{2}x = 2tan\:x\\\\ \tt\implies cos^{2}x - sin^{2}x = 2tan\:x$

$\tt\implies \: \frac{ {cos}^{2} x - { \sin}^{2}x }{2tanx} = 1$

Answer 2

Answer:

$\huge {\purple {\fbox {\bigstar {\mathfrak {\red {hello\:mate}}}}}}$

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$\huge {\purple {\fbox {\bigstar {\mathfrak {\red {ur\: question}}}}}}$

$\huge {\mathfrak {\purple {Q.}}}}$sin²x=1-tan2x genra formulat

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$\huge {\purple {\fbox {\bigstar {\mathfrak {\red {ur\: ANSWER✓}}}}}}$

$\tt\implies sin^2x = 1 - tan2x$

$\purple{\text{we \:know\: that\: ,}}$

$\tt\implies tan2x = \frac{2tanx}{1 - {tan}^{2}x }$

$\red{\text{now}}$

$\tt\implies cos^2x ( 1 - tan^2x ) = 2tanx$

$\tt\implies cos^2x - cos^2x tan^2x = 2tanx$

$\tt\implies cos^2x - sin^2x = 2tanx$

$\tt\implies \frac{ {cos}^{2} x - { \sin}^{2}x }{2tanx}$

$\sf{\blue{\fbox{\purple{\bigstar{\mathbf{\red{answer=\:\frac{ {cos}^{2} x - { \sin}^{2}x }{2tanx}}}}}}}}$

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