If 4\x \sin^{2}60^\circ + 3 \sec^{2}45^\circ = 7\x \sec60^\circ – 9 \tan45^\circ \cosec30^\circ , then, find the value of \x .

Haryana Police Constable Model Paper 8

Question : 56 of 100

Marks: +1, -0

4\x \sin^{2}60° + 3 \sec^{2}45° = 7\x \sec60° – 9 \tan45° \cosec30°

, then, find the value of

\x

Solution:

⇒4\x×(√3/2)^{2}+3(√2)^{2}=7\x×(2)−9(1)(2)

⇒ 3\x + 6 = 14\x -18

⇒ 14\x -3\x = 6 + 18

⇒ 11\x = 24

⇒ \x = 24/11

$∴ 24/11$

Go to Question:

Prev Question

Next Question