Answer:
[tex]4 \cot {}^{2} (45) - 3 \tan {}^{2} (60) + 2 \sec {}^{2} (60) = 3 \\ [/tex]
[tex]LHS = 4 \cot {}^{2} (45) - 3 \tan {}^{2} (60) + 2 \sec {}^{2} (60) \\[/tex]
let us first take a look at the values of the trigonometric ratios given in the question so that we get quite clear about what is to be done.
here ,
[tex] \cot(45) = 1 \\ \\ \tan(60) = \sqrt{3} \\ \\ \sec(60) = 2[/tex]
now ,
we just have to plug in the values considering certain other things given in the question and we're done!
so let's start ~
[tex]4(1) {}^{2} - 3( \sqrt{3} ) {}^{2} + 2(2) {}^{2} \\ \\ \dashrightarrow \: 4(1) - 3(3) + 2(4) \\ \\ \dashrightarrow \: 4 - 9 + 8 \\\\ \dashrightarrow \: 12 - 9 \\ \\\dashrightarrow \: 3 = RHS[/tex]
hence , proved ~
hope helpful :D
