Question

find the valhe of sin^2threta +1/1+tan^2threta

Answer 1

sin²@ + 1/(1+tan²@),

sin²@ + 1/sec²@,

sin²@ + cos²@,

1

Answer 2

$\frac{ { \sin }^{2}x + 1}{1 + { \tan }^{2}x }$
as tanx = sinx/cosx
our function becomes
$\frac{ {sin}^{2}x + 1 }{ 1 + \frac{ {sin}^{2}x }{ {cos}^{2}x } }$
which can be further written as
${cos}^{2} x \: ( {sin}^{2} x \: + 1)$
as in denominator (cosx)^2 + (sinx)^2 = 1
${cos}^{2} x \: = 1 - {sin}^{2} x$
the function becomes
$(1 - {sin}^{2} x)(1 + {sin}^{2}x )$
applying formula
$(a - b)(a + b) = {a}^{2} - {b}^{2}$

$1 - {sin}^{4} x$