Question

$\frac{ { \cos}^{2} (45 + x) + { \cos}^{2} (45 - x)}{\tan(60 + x) \tan(30 + x)} = 1$
please prove it
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Answer 1

Step-by-step explanation:

as we know

cosA=sin(90-A)

sinA=cos(90-A)

tanA=cot(90-A)

n

tanA×cotA=1

$\frac{ {cos}^{2}(45 + x) + {cos}^{2} (45 - x) }{tan(60 + x)(tan(30 + x)} \\ = \frac{ {cos²}(45+x)+sin²(90-(45-x)) }{tan(60+x)(cot(90-(30+x))} \\ = \frac{cos²(45+x)+sin²(45+x)}{tan(60+x)×cot(60+x)} \\ = \frac{1}{1 = 1}$