Question

Minimum value of Sin, 0 + cos O Hando
+ cot20+ Sec²o - cosec O.R.
0​

Answer 1

Answer:

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Step-by-step explanation:

ANSWER

f(x)=sin

2

θ+cosec

2

θ+2+cos

2

θ+sec

2

θ+2

f(x)=5+cosec

2

θ+sec

2

θ

f

′

(x)=0=−2cosec

2

θcotθ+2sec

2

θtanθ

2cosec

2

θcotθ=2sec

2

θtanθ

tan

4

θ=1

θ=

4

π

f(x)=5+cosec

2

θ+sec

2

θ=5+2+2=9

Reason:

consider, sin2θ=2cosθsinθ

Max value of x for both sin and cos together is 45

o

which is 1/

2

∴sin2θ=2×

2

1

×

2

1

=1

Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1