Question

Prove that %20%3D%203" id="TexFormula1" title="4cot45 ^{2} - 3tan ^{2} 60 + 2sec ^{2}60 = 3" alt="4cot45 ^{2} - 3tan ^{2} 60 + 2sec ^{2}60 = 3" align="absmiddle" class="latex-formula">
help me​

Answer 1

Answer:

$4 \cot {}^{2} (45) - 3 \tan {}^{2} (60) + 2 \sec {}^{2} (60) = 3$

$LHS = 4 \cot {}^{2} (45) - 3 \tan {}^{2} (60) + 2 \sec {}^{2} (60)$

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 ,

$\cot(45) = 1 \\ \\ \tan(60) = \sqrt{3} \\ \\ \sec(60) = 2$

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 ~

$4(1) {}^{2} - 3( \sqrt{3} ) {}^{2} + 2(2) {}^{2} \\ \\ \dashrightarrow \: 4(1) - 3(3) + 2(4) \\ \\ \dashrightarrow \: 4 - 9 + 8 \\\\ \dashrightarrow \: 12 - 9 \\ \\\dashrightarrow \: 3 = RHS$

hence , proved ~

hope helpful :D