Question

cos²Ø/1-tanØ+sin³ø/sinø-cosØ=1+sinØ× cosø​

Answer 1

Step-by-step explanation:

I hope you were able to answer this question correctly

Answer 2

$\huge {\bold {HELLO..!,}}$

$\frac{{cos}^{2} \: x}{ 1 - {tan}^{2}x } + \frac{ {sin}^{3} x}{sin \: x \: - cos \: x} = 1 - sin \: x.cos \: x$

$= > \frac{{cos}^{2} \: x}{ 1 - {tan}^{2}x } + \frac{ {sin}^{3} x}{sin \: x \: - cos \: x} \: \\ \\ = > \frac{{cos}^{2} \: x}{ 1 - \frac{ { sin }^{2} \: x }{ { cos }^{2} \: x} } + \frac{ {sin}^{3} x}{sin \: x \: - cos \: x} \: \\ \\ = > \frac{{cos}^{3} \: x}{ { cos }^{2} x - { sin }^{2}x } + \frac{ {sin}^{3} x}{sin \: x \: - cos \: x} \: \\ \\ = > \frac{ { cos }^{3} x - { sin }^{3} x}{ cos \: x - sin \: x } \\ \\ = > \frac{(cos \: x - sin \: x)( { cos }^{2} \: x + cos \: x.sin x + { sin }^{2} \: x )}{( cos \: x - sin \: x)} \\ \\ = > { cos}^{2} + { sin}^{2} + sin \: x .cos \: x \\ \\ = > 1 + sin \: x .cos \: x$

$Here, \:LHS = RHS.$

$\huge {\bold {Hope\: It \: Helps }}$