Question

PLZ GIVE RELAVENT ANSWERS ONLY...............................
If x = log [cot (π/4 + θ)], then consider the following statements
when θ∈( -π/4 , π/4 )
I. cosh x = sec 2θ II. Sinh x = – tan 2θ
1) I is true and II is false 2) I is false and II is true 3) Both I and II are true 4) Both I and II are false

Answer 1

Solution:

Solution:-First let us find the value of coshx

as we know

coshx={e^x+e^-x)/2=(e^2x+1)/(2e^x)

analysing the log term we get

e^x=cot(π/4 +∅)

now simply we will put the value in the formula of coshx

coshx={cot^2(π/4+∅)+1}/{2×cot(π/4+∅)}

coshx={cos^2(π/4 +∅)+sin^2(π/4 +∅)}/{sin^2(π/4 +∅)(2×cos(π/4+∅)/sin(π/4+∅)}

coshx=1/{sin(π/4+∅)×2×cos(π/4+∅)}

coshx=1/{(sin2(π/4+∅)}

coshx=1/{sin(π/2+2∅)}

coshx=1/cos2∅

coshx=sec2∅

∅Hence I statement is correct

Let us analyse II statement

As we know that,

sinhx={(e^x-e^-x)}/2={(e^x-1/e^x)}/2=(e^2x-1)/(2e^x)

now.putting the value of

e^x=cot(π/4+∅)

sinhx={cot^2(π/4+∅)-1}/{2×cot(π/4+∅)}

now convert it into tan term

sinhx={1/tan^2(π/4+∅) -1}/{2×1/tan(π/4+∅)}

sinhx={tan(π/4 +∅)(1-tan^2(π/4 +∅)}/{2tan^2(π/4+∅)}

sinhx={1-tan^2(π/4 +∅)}/{2tan(π/4+∅)}.....i)

from the formula of Tan2∅

we know,Tan2∅=(2tan∅)/(1-tan^2∅)

2tan∅=(tan2∅)(1-tan^2∅)

Similarly,

2Tan(π/4+∅)=Tan2(π/4+∅)×(1-tan^2(π/4 +∅)

putting this value in i) we get

sinhx={1-tan^2(π/4+∅)}/{(tan2(π/4+∅))(1-tan^2(π/4+∅)}

1-tan^2(π/4+∅),will be cancelled out from numerator and denominator

then we get

sinhx=1/tan2(π/4+∅)=1/(tan(π/2+2∅)

sinhx=1/{tan(π/2+2∅)}

sinhx=1/-cot2∅

sinhx=-tan2∅

Hence II statement is also right

So,Option3) is the right one