IIT JEE , AIEEE Forum - Mathematics , Trignometry

sathya_crazyteen

1.pt (sin x+cosec x)²+(cos x+ sec x)²>= 9

2.eliminate

between cosec

-sin

=m; sec

- cos

=n

3.if sin x+sin y=a, cos x+cos y=b and tan x+ tan y=c,st ((a²+b²)²-4a²)c= 8ab

abiiizer

3 one is simple

pink_ele

(sinx+cosecx)² +(cosx+secx)²=sin ²x+cos²x +2+2+cosec²x +sec²x

=5+(secxcosecx)²

for MINIMUM VALUE X=

/4

MINIMUM=5+(

2)²=9

^{[ ]}

nunoxic

1/sin@ - sin@ = m
1 - sin^2 @/sin@ = m
cos^2 @ /sin @ = m
similarly
sin^2 @ / cos @ = n

joyfrancis

(sinx+cosecx)^2 + (cosx + secx)^2

expand

sin²x+cosec²x+2+cos²x+sec²x+2

5+cosec²x+sec²x

5+(1/sinxcosx)

5+(1/{(2/2)sinxsosx}²)

5+4/sin2x

for minimum vale sin2x should be maximium i.e 1

therefore

5+4 = 9

Therefore

(sin x+cosec x)²+(cos x+ sec x)²>= 9

Avinash_Bhat

1)

(sin x + cosec x) + (cos x + sec x)²

= sin²x +cosec²x + 2 + cos²x + sec²x + 2

= 5 + cosec²x + sec²x

= 5 + tan²x + cot²x + 2

>= 9 (minimum value of tan²x + cot²x =2)

2)

cosec

- sin

=m ie, (1 - sin²

) / sin

= cos²

/ sin

=m

Similarily: sin²

/ cos

= n

Now m²n = cos³

and mn² = sin³

ie, sin²

+ cos²

= (mn²)^2/3 + (m²n)^2/3 =1

SO : m ^2/3n^2/3 (m^2/3+ n^2/3 ) = 1

I will be posting the last one later.....

Avinash_Bhat

The last one is simple yaar.., just substitute the values of a, b, and c in ((a² + b²)² - 4a²)c, and on simplification you will get it as 8ab.

Ask & Discuss Questions with Community & Experts