Question

if tan theta +1/tan theta =2 ;show that :tan²theta+1/ten² theta=2​

Answer 1

Answer:

$\tan( \alpha ) + \frac{1}{ \tan( \alpha ) } = 2 \\ ( \tan( \alpha ) + \frac{1}{ \tan( \alpha ) } ) {}^{2} = \tan {}^{2} ( \alpha ) + \frac{1}{ \tan {}^{2} ( \alpha ) } + 2 \\ 2 ^{2} = tan {}^{2} ( \alpha ) + \frac{1}{ \tan {}^{2} ( \alpha ) } + 2 \\ tan {}^{2} ( \alpha ) + \frac{1}{ \tan {}^{2} ( \alpha ) } = 2 {}^{2} - 2 = 4 - 2 = 2 \\ ie \: \: \: \: tan {}^{2} ( \alpha ) + \frac{1}{ \tan {}^{2} ( \alpha ) } = 2$

Answer 2

Step-by-step explanation:

Given that,

$\sf{tan \: x + \frac{1}{tan \: x} = 2 ............(1)}$

To prove:

$\sf{tan {}^{2}x + \frac{1}{tan {}^{2} x} = 2 }$

Squaring (1) on both sides,

$\sf{(tan \: x + \: \frac{1}{tan \: x} ) {}^{2} = 2 {}^{2} } \\ \\ \implies \: \sf{tan {}^{2}x + 2tan \: x. \frac{1}{tan \: x} + \frac{1}{ tan {}^{2}x } = 4} \\ \\ \implies \: \sf{tan {}^{2}x + \frac{1}{tan {}^{2}x } = 4 - 2 } \\ \\ \implies \: \boxed{ \boxed{ \sf{tan {}^{2}x + \frac{1}{tan {}^{2}x} = 2 }}}$

Hence,proved

•Some basic trigonometric identities:

sin²x + cos²x=1

cosec²x-cot²x=1

sec²x-tan²x=1

•Trigonometric relations:

sin x=1/cosec x

cos x=1/sec x

tan x=1/cot x

Note,I have used x instead of theta