sinh[log(2+√5]+cosh[log(2+√3)]=

Correct answer: (A) 4. We know sinhx=(e^x-e^(-x))/2 & coshx=(e^x+e^(-x))/2. So, sinh[log(2+√5)]+cosh[log(2+√3)]=(e^(log2+5)-e^(-log2+5))/2+(e^(log2+3)+e^(-log2+3))/2 ⇒sinhlog2+√5+coshlog2+√3=((2+√5)-1(2+√5))/2+((2+√3)+1(2+√3))/2 ⇒sinh[log2+√5]+cosh[log2+√3]=4.

Q64 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

$\sin h [\log ((2 + \sqrt{5})] + \cos h [\log (2 + \sqrt{3})] =$

Correct answer
A
$4$
B
$3$
C
$2$
D
$1$

Solution

We know

$\sin h (x) = \frac{e^{x} - e^{- x}}{2} & \cos h (x) = \frac{e^{x} + e^{- x}}{2}$ .

So,

$\sin h [\log (2 + \sqrt{5})] + \cos h [\log (2 + \sqrt{3})] = \frac{e^{\log (2 + \sqrt{5})} - e^{- \log (2 + \sqrt{5})}}{2} + \frac{e^{\log (2 + \sqrt{3})} + e^{- \log (2 + \sqrt{3})}}{2}$

$\Rightarrow \sin h [\log (2 + \sqrt{5})] + \cos h [\log (2 + \sqrt{3})] = \frac{(2 + \sqrt{5}) - \frac{1}{(2 + \sqrt{5})}}{2} + \frac{(2 + \sqrt{3}) + \frac{1}{(2 + \sqrt{3})}}{2}$

$\Rightarrow \sin h [\log (2 + \sqrt{5})] + \cos h [\log (2 + \sqrt{3})] = 4$ .

Hence, the correct answer is option: 'A'.

sinh[log(2+5]+cosh[log(2+3)]=

Solution

Relevant Questions

$\sin h [\log ((2 + \sqrt{5})] + \cos h [\log (2 + \sqrt{3})] =$