Question

show that --->
cos^2(45°+ø)+cos^2(45°-ø)/tan(60°+ø)tan(30°-ø) = 1​

Answer 1

Answer:

we know that tanπ/2= tan{(π/3+x)+(π/6-x)}[here i am writingФ as x]

we know the formula tan(x+y)=tanx+tany/1-tanxtany

since tanπ/2=1/0

when we substitute that we get

tanπ/3+xtanπ/6-x=1

also  

cos^2π/4+x=sin^2π/2-(π/4+x)=sin^2(π/4-x)

thus the numerator becomes of the form

sin^2x+cos^2x which is 1 (by identity)

thus both numerator and denominator became 1

hence proved

hope u understood the answer

please understand the hard work i did to write this much and considering it please mark my answer as the brainliest one

Answer 2

Answer:

$\frac{ \cos {}^{2} (45 + \alpha ) + \cos {}^{2} (45 - \alpha )}{ \tan(60 + \alpha ) \tan(30 - \alpha ) }$

$= \frac{ {( \cos(45) \cos( \alpha ) - \sin(45) \sin( \alpha ) )}^{2} + {(\cos(45) \cos( \alpha ) + \sin(45) \sin( \alpha ) )}^{2} }{ \cot(90 - (60 + \alpha )) \tan(30 - \alpha ) }$

$= \frac{( \frac{1}{ \sqrt{2} } \cos( \alpha ) - \frac{1}{ \sqrt{2} } \sin( \alpha )) {}^{2} + \frac{1}{ \sqrt{2} } \cos( \alpha ) + \frac{1}{ \sqrt{2} } \sin( \alpha )) {}^{2} }{ \cot(30 - \alpha ) \tan( 30 - \alpha ) }$

$= \frac{1}{2} \frac{ {( \cos( \alpha ) - \sin( \alpha ) )}^{2} + {( \cos( \alpha ) + \sin( \alpha ) )}^{2}}{1}$

$now \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

Therefore

$\frac{1}{2} \times 2( \cos { }^{2} ( \alpha ) + \sin {}^{2} ( \alpha ))$

$= \frac{1}{2} \times 2 \times 1$

$= 1$

Hence proved