Question

find the value of 1 - tan square thetha upon 1 + tan square thetha​

Answer 1

Answer:

Step-by-step explanation:

(remembering that  

tan

x

=

sin

x

cos

x

and  

sin

2

x

+

cos

2

x

=

1

),

the second member becomes:

1

−

sin

2

x

cos

2

x

1

+

sin

2

x

cos

2

x

=

cos

2

x

−

sin

2

x

cos

2

x

cos

2

x

+

sin

2

x

cos

2

x

=

=

(

cos

2

x

−

sin

2

x

cos

2

x

)

⋅

cos

2

x

cos

2

x

+

sin

2

x

=

=

cos

2

x

−

sin

2

x

,

that is the development of the formula of  

cos

2

x

.

Answer 2

Answer:-

→$\bf{ {cos}^{2} \theta - {sin}^{2} \theta = cos \: 2 \theta}$

Explanation:-

$\implies \bf{\frac{1 - {tan}^{2} \theta }{1 + {tan}^{2} \theta} } \\ \\ \mathfrak{ \because \: {sec}^{2} \theta - {tan}^{2} \theta = 1} \\ \mathfrak{ \therefore \: 1 + {tan}^{2} \theta = {sec}^{2} \theta} \\ \\ </u><u>H</u><u>ence</u><u>,</u><u> \\ \\ \implies \: \bf{\frac{1 - {tan}^{2} \theta }{ {sec}^{2} \theta } } \\ \\ \implies \: \bf{ \frac{1}{ {sec}^{2} \theta} - \frac{ {tan}^{2} \theta }{ {sec}^{2} \theta } } \\ \\ \bf{ \implies \: {cos}^{2} \theta - {tan}^{2} \theta. {cos}^{2} \theta} \\ \\ \bf{ \implies \: {cos}^{2} \theta - \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta} . {cos}^{2} \theta} \\ \\ \bf{ \implies \: {cos}^{2} \theta - {sin}^{2} \theta} \\ \\ \bf{ \implies \: cos \: 2 \theta}$

Hope it helps you.