Question

find the value of:
(3 tan 41/cot 49)^2 - (sin35sec55/tan10tan20tan60tan70tan80)^2

Answer 1

Answer:

$\frac{26}{3}$

Step-by-step explanation:

Step 1 :-

cot 49 = cot (90 - 41) = tan 41

∴ ( 3 × $\frac{tan 41}{cot 49}$)²

= ( 3 × $\frac{tan 41}{tan 41}$)²

= ( 3 × 1 )²

= 9

Step 2 :-

sec 55 = sec (90 - 35) = cosec 35

∴ ( sin 35 × sec 55 ) = ( sin 35 × cosec 35 ) = 1

Step 3 :-

tan 80 = tan (90 - 10) = cot 10

tan 70 = tan (90 - 20) = cot 20

∴ ( tan 10 × tan 20 × tan 60 × tan 70 × tan 80 )

= ( tan 10 × cot 10 × tan 20 × cot 20 × tan 60 )

= ( 1 × 1 × tan 60 )

= $\sqrt{3}$

Hence,

( 3 × $\frac{tan 41}{cot 49}$)² - ( 3 × $\frac{sin 35 × sec 55}{tan 10 × tan 20 × tan 60 × tan 70 × tan 80}$)²

= 9 - ($\frac{1}{\sqrt{3}}$)² [Using values from step 1, 2 and 3]

= 9 - $\frac{1}{3}$

= $\frac{26}{3}$

Answer 2

Answer:

26/3

Step-by-step explanation:

Given find the value of:

(3 tan 41/cot 49)^2 - (sin35 sec55 / tan10 tan20 tan60 tan70 tan80)^2

(3 tan 41 / cot (90 - 41) -  sin 35 sec(90 - 35) / tan 10. tan 20. tan 60 . tan(90 - 20) tan(90 - 10)

(3 tan 41 / tan 41)^2 - (sin 35 sec(90 - 35) / tan 10 tan 20 tan 60 tan(90 - 20)tan(90 - 10))^2

(3 x 1)^2 - (sin 35 cosec 35 / tan 10. cot 10. tan 20. cot 20. tan 60)^2

9 - (1 / 1 x 1 x √3)^2      (because cosec 35 = 1/sin35)

9 - 1 / 3

27 - 1 / 3

 26 / 3