Question

tan0/1-cot0+cot0/1-tan0=1+tan0+cot0​

Answer 1

$=1+tan\Theta +cot\Theta$       LHS=RHS

Step-by-step explanation;

  Given,

$tan\Theta /\left ( 1-cot\Theta \right )+cot\Theta /\left ( 1-tan\Theta \right )$ =$=tan\Theta +cot\Theta +1$

we solve LHS

$=sin\Theta/(cos\Theta ( 1-cos\Theta /sin\Theta \right ))+cos\Theta /(sin\Theta ( 1-sin\Theta /cos\Theta \right ))$

$=\left ( sin\Theta \right )^{2}/\left ( cos\Theta \left ( sin\Theta -cos\Theta \right ) \right )+\left ( cos\Theta \right )^{2}/\left ( sin\Theta \left ( cos\Theta -sin\Theta \right ) \right )$

$=\left ( sin\Theta \right )^{2}/\left ( cos\Theta \left ( sin\Theta -cos\Theta \right ) \right )-\left ( cos\Theta \right )^{2}/\left ( sin\Theta \left ( sin\Theta -cos\Theta \right ) \right )$

$=1/\left ( sin\Theta -cos\Theta \right )\left ( \left ( sin\Theta \right )^{2}/cos\Theta -\left ( cos\Theta \right )^{2} /sin\Theta \right )$

$=1/\left ( sin\Theta -cos\Theta \right )\left ( \left ( sin\Theta \right )^{3} -\left ( cos\Theta \right )^{3}\right )/\left ( sin\Theta \times cos\Theta \right )$

use formula $a^{3}-b^{3}=\left ( a-b \right )^{3}+3ab\left ( a-b \right )$  

$=\left ( sin\Theta -cos\Theta \right )^{3}+3sin\Theta \times cos\Theta \left ( sin\Theta -cos\Theta \right )/\left ( \left ( sin\Theta -cos\Theta \right )\left ( sin\Theta \times cos\Theta \right ) \right )$

taking $(sin\Theta -cos\Theta )$ as common and cancel numerator and denominator by $(sin\Theta -cos\Theta )$

$=(\left ( sin\Theta -cos\Theta \right )^{2}+3\times sin\Theta \times cos\Theta)/\left ( sin\Theta \times cos\Theta \right )$

$=\left ( \left ( sin\Theta \right )^{2}+\left ( cos\Theta \right )^{2} -2\times sin\Theta \times cos\Theta +3\times sin\Theta \times cos\Theta \right )/\left ( sin\Theta cos\Theta \right )$

                                                                                                                                                                                                                                                                                                $=\left ( sin\Theta \right )^{2}/\left ( sin\Theta \times cos\Theta \right )+\left ( cos\Theta \right )^{2}/\left ( sin\Theta \times cos\Theta \right )+\left ( sin\Theta \times cos\Theta \right )/\left ( sin\Theta \times cos\Theta \right )$        use formula $sin\Theta /cos\Theta =tan\Theta$            $cos\Theta /sin\Theta =cot\Theta$

$=tan\Theta +cot\Theta +1$

$=1+tan\Theta +cot\Theta$

LHS=RHS   hence proved

Answer 2

Answer:

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