Question

Solve it⤵

tan(65-θ) - cot(25+θ) = tan{90-(25+θ)}

$\Rightarrow$ Also tell value of tan{90-(25+θ)}​

Answer 1

Therefore, this should be calculated by = { As we know, Tan θ = Cot ( 90 - θ ) , so :- }We get, Tan [ 90 - ( 65 + θ ) ] - Cot ( 25 - θ ) Cot [ 25 - θ ] - Cot ( 25 - θ)= 0.

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Answer 2

$\large\boxed{ \underline{\underline{ \blue{ \sf{question : }}}}}$

$\sf{tan (65 \degree - \theta) - cot(25 \degree + \theta) = tan(90 \degree - (25 \degree + \theta))}$

$\large\boxed{ \underline{\underline{ \blue{ \sf{formulae : }}}}}$

$\sf{tan \: \theta =cot(90 \degree - \theta) } \\ \sf{ cot \: \theta = tan (90 \degree - \theta)}$

$\large\boxed{ \underline{\underline{ \blue{ \sf{solution : }}}}}$

$\sf{tan (65 \degree - \theta) - cot(25 \degree + \theta) = tan(90 \degree - (25 \degree + \theta))} \\ \sf{tan(65 \degree - \theta) - tan(90 \degree - (25 \degree + \theta)} = tan(90 \degree - ( 25 \degree + \theta)) \\ \sf{tan(65 \degree - \theta) - tan(90 \degree - 25 \degree - \theta)} = tan(90 \degree - ( 25 \degree + \theta)) \\ \sf{tan(65 \degree - \theta) - tan(65 \degree - \theta)} = tan(90 \degree - ( 25 \degree + \theta)) \\ \sf{ \cancel{tan(65 \degree - \theta)} - \cancel{tan(65 \degree - \theta)}} = tan(90 \degree - ( 25 \degree + \theta)) \\ \sf{0 \: \: \: \: \: \: = \: \: tan(90 \degree - ( 25 \degree + \theta))} \\ \boxed{ \sf{tan(90 \degree - ( 25 \degree + \theta)) = 0}}$