IIT JEE , AIEEE Forum - Mathematics , Trignometry

ankitseth

(cot 54/tan 36)+ (tan 20+cot 70)

i cud simplify it till 1 +2cot70....(everyone can do dat!!!!)

a detailed solution wud b appreciated!!!!!!

10904him

Thogh I could not find the answer by trigo----
But on simplification I got --
= 2
---------------------------
cos45 + sin 25
since for small angle(rad) sin x =x
Thus sin 25 =25xpi/180 = 0.436
Thus =2/(0.707+0.436)
=1.75
=(nearly) (3)^1/2 Ans
We can use this in case if we need this simplification in physics etc but not in Maths
Still trying to solve it.

sanchay_1991

(cot 54/tan 36) + (tan 20 + cot 70)

=(tan 36/tan 36) + (sin 20/cos 20 + cos 70/sin 70)

=1+ (sin 20*sin 70 +cos 20*cos 70/cos20*sin 70)

=1+ (cos 70-20/cos 20* sin 70)

=1+ (2cos 50/ 2sin 70* cos 20)

=1+ (2cos 50/sin 45+ sin 25)

=1+ (2

2 cos 50/1 +

2 sin25)

= 1+ (2

2 (1 - 2 sin²25) / 1 +

2sin25)

=1+ {2

2 (1 -

2 sin 25)

= 1 +{ 2

2 - 4 sin 25)

= 1+ {2

2 - 4

1 -cos 45/2}

now put the value of cos45 n get the ans

you will get ans as around root 5 or by putting value orf root 2 as 1.414you will get as 2.29769

do rate me if u get the ans

ankitseth

i appreciate it sanchay............but da answer shud come 2......10904him's answer is correct......not too sure abt da method!!!!

Asmita

4 the ans. 2 b =2. tan 20 must b =1/2.

tan3(20)=tan 60=rt3

i.e.3tan20-tan^3 20/1-3tan^2 20=rt 3

which is not true if u replace tan 20 by 1/2.

sanchay_1991

can u plesse point my mistake

so i can improve it

pls reply fast

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