Question

Answer this Question. The question is in the above photo

Answer 1

$\huge \purple{\mathfrak{solution}}$

To find the value of :

$\boxed{ \bold{ \frac{ {tan}^{2} \theta - {sec}^{2} \theta }{ {cot}^{2} \theta - {cosec}^{2} \theta } }}$

Take minus common from both numerator and denominator.

$\boxed{ \red\leadsto} \bold{\frac{ - ( {sec}^{2} \theta - {tan}^{2} \theta) }{ - ( {cosec}^{2} \theta - {cot}^{2} \theta) } } \\ \\ \boxed{ \red \leadsto} \frac{ - 1}{ - 1} \\ \\ \boxed{ \red \leadsto} \: 1$

Hence , the value is

$\huge \boxed{ \green{ \boxed{1}}}$

Formula used :

$\boxed{ \bold{\blue{ {sec}^{2} \theta - {tan}^{2} \theta = 1}}}$

$\boxed{ \bold{ \pink{{cosec}^{2} \theta - {cot}^{2} \theta = 1}}}$

Answer 2

Question :--

$\frac{ {tan}^{2} a - {sec}^{2}a }{ {cot}^{2} a - {cosec}^{2}a }$

Formula used :---

• sec²a - tan²a = 1
• cosec²a - cot²a = 1

Solution :------

Taking (-1) common from numerator and denominator of Question we get,

$\frac{ ( \cancel - 1)({sec}^{2} a - {tan}^{2}a) }{( \cancel - 1)( {cosec}^{2} a - {cot}^{2}a )} \\ \\ = \frac{ ({sec}^{2} a - {tan}^{2}a) }{( {cosec}^{2} a - {cot}^{2}a )} \\ \\ = \frac{1}{1} = 1$

(Hope it helps you)