Question

cos95° + cos 25° = 1/√2(cos10° + sin 10°)​

Answer 1

$\textbf{To prove:}$

$cos\,95^{\circ}+cos\,25^{\circ}=\dfrac{1}{\sqrt{2}}[cos\,10^{\circ}+sin\,10^{\circ}]$

$\textbf{Solution:}$

$\text{Consider,}$

$cos\,95^{\circ}+cos\,25^{\circ}$

$\text{Using the formula}$

$\boxed{\bf\,cosC+cosD=2\,cos(\frac{C+D}{2})\,\,cos(\frac{C-D}{2})}$

$=2\,cos(\frac{95^{\circ}+25^{\circ}}{2})\,cos(\frac{95^{\circ}-25^{\circ}}{2})$

$=2\,cos(\frac{120^{\circ}}{2})\,cos(\frac{70^{\circ}}{2})$

$=2\,cos\,60^{\circ}\,cos\,35^{\circ}$

$=2(\frac{1}{2})cos\,35^{\circ}$

$=cos\,35^{\circ}$

$=cos(45^{\circ}-10^{\circ})$

$\text{Using the formula}$

$\boxed{\bf\,cos(A-B)=cosA\,cosB+sinA\,sinB}$

$=cos\,45^{\circ}\,cos\,10^{\circ}+sin\,45^{\circ}\,sin\,10^{\circ}$

$=(\frac{1}{\sqrt{2}})cos\,10^{\circ}+(\frac{1}{\sqrt{2}})sin\,10^{\circ}$

$=\frac{1}{\sqrt{2}}[cos\,10^{\circ}+sin\,10^{\circ}]$

$\therefore\bf\,cos\,95^{\circ}+cos\,25^{\circ}=\dfrac{1}{\sqrt{2}}[cos\,10^{\circ}+sin\,10^{\circ}]$

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Answer 2

HI DEAR SHRESTA HERE!!!PLS SEE THE ANSWER !!!!

USING FORMULAS:-

COS C+COS D  = 2COS(C+D/2).COS(C-D/2)

COS(A-B)  =  COS A.COS B  +  SIN A.SIN B

LHS:-

=2cos(95+25/2)cos(95-25/2)

=2cos(120/2)cos(70/2)

=2cos60cos35

=2*1/2*cos 35

=cos35

=cos(45-10)

=cos45cos10 +sin45sin10

=1/√ 2 .cos10 +  1/√ 2. sin10

=1/√ 2(cos10  +  sin10)

=RHS

PLS THANK MY ANSWER ....!!!!!!!!!

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