Question

prove that 2(cos^2 45+tan^2 60)-6 (sin^2 45 -tan^2 30)=6

Answer 1

hopes i helped...........

Answer 2

Answer:

$2( cos^{2}45\degree+tan^{2}60\degree)-6(sin^{2}45\degree-tan^{2}30\degree)=6$

Step-by-step explanation:

$LHS = 2( cos^{2}45\degree+tan^{2}60\degree)-6(sin^{2}45\degree-tan^{2}30\degree)$

$= 2[\left( \frac{1}{\sqrt{2}}\right)^{2}+(\sqrt{3})^{2}]-6[\left( \frac{1}{\sqrt{2}}\right)^{2}-\left(\frac{1}{\sqrt{3}}\right)^{2}]$

$= 2\left( \frac{1}{2}+3\right) - 6 \left(\frac{1}{2}-\frac{1}{3}\right)\\=2\left( \frac{1+6}{2}\right)-6\left(\frac{3-2}{6}\right)\\=2\times \frac{7}{2}-6\times \frac{1}{6}\\=7 - 1\\=6\\=RHS$

Therefore,

$2( cos^{2}45\degree+tan^{2}60\degree)-6(sin^{2}45\degree-tan^{2}30\degree)=6$

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