Question

If tan 2tita = cot(tita +6°) then tita =_________

Answer 1

AnsWer :

theta = 28°

SolutioN :

$\tt \: if \: tan2 \theta = cot( \theta + 6)$

We know that.

$\tt \dagger \: \: \: \: \: tan( 90 - \theta ) = cot \theta.$

Substitute the value of tan theta as place of cot theta.

$\tt : \implies tan2 \theta = tan(90 - ( \theta + 6))$

$\tt : \implies 2 \theta =(90 - \theta - 6)$

$\tt : \implies 3\theta = 84.$

$\tt : \implies \theta = \dfrac{ \cancel{84}}{ \cancel3}.$

$\tt : \implies \theta =28.$

Therefore, the value of theta is 28°.

Some More FormulA :

$\tt \dagger \: \: \: \: \: sin( 90 - \theta ) = cos \theta.$

$\tt \dagger \: \: \: \: \: cos( 90 - \theta ) = sin \theta.$

$\tt \dagger \: \: \: \: \: tan( 90 - \theta ) = cot \theta.$

$\tt \dagger \: \: \: \: \: cot( 90 - \theta ) = tan \theta.$

Answer 2

Question :–

▪︎ If $\: \: { \bold{ \tan(2 \theta) = \cot( \theta + {6}^{ \circ}) }} \: \:$ then $\: \: { \bold{ \theta = ? }} \: \:$

ANSWER :–

$\\ \longrightarrow \large { \pink{ \boxed{ \bold{ \theta = {28}^{ \circ} }}}} \: \:$

EXPLANATION :–

$\\ \bigstar \: \: \: { \red{ \boxed{ \bold{ \tan( \theta) = \cot({90}^{ \circ} - \theta ) }}}} \: \:$

• So that –

$\\ \implies \: { \bold{ \cot( {90} ^{ \circ} - 2 \theta) = \cot( \theta + {6}^{ \circ}) }} \: \:$

$\\ \implies \: { \bold{ {90}^{ \circ} - 2 \theta = \theta + {6}^{ \circ} }} \: \:$

$\\ \implies \: { \bold{ {90}^{ \circ} - {6}^{ \circ} = \theta + 2 \theta }} \: \:$

$\\ \implies \: { \bold{ {84}^{ \circ} = 3 \theta }} \: \:$

$\\ \implies \: { \bold{ \theta = \dfrac{ \: {84}^{ \circ}}{3} }} \: \:$

$\\ \implies \large { \pink{ \boxed{ \bold{ \theta = {28}^{ \circ} }}}} \: \:$

Other important formulas :–

$\\ \bigstar \: \: \: { \red{{ \bold{ \sin( \theta) = \cos({90}^{ \circ} - \theta ) }}}} \: \:$

$\\ \bigstar \: \: \: { \red{{ \bold{ \cos( \theta) = \sin({90}^{ \circ} - \theta ) }}}} \: \:$

$\\ \bigstar \: \: \: { \red{{ \bold{ \tan( \theta) = \cot({90}^{ \circ} - \theta ) }}}} \: \:$

$\\ \bigstar \: \: \: { \red{{ \bold{ \cot( \theta) = \tan({90}^{ \circ} - \theta ) }}}} \: \:$

$\\ \bigstar \: \: \: { \red{{ \bold{ \sec( \theta) = cosec({90}^{ \circ} - \theta ) }}}} \: \:$

$\\ \bigstar \: \: \: { \red{{ \bold{ cosec( \theta) = \sec({90}^{ \circ} - \theta ) }}}} \: \:$