Question

Please Give the correct solution of this answer.

Answer 1

Given:

$\longrightarrow\rm{cos\:\theta+cos\:\phi=a}$

$\longrightarrow\rm{sin\:\theta+sin\:\phi=b}$

To Find:

Value of cosθ.cosΦ

Solution:

We know that,

$\rightarrow\rm{sinA+sinB=2sin\bigg(\dfrac{A+B}{2}\bigg)cos\bigg(\dfrac{A-B}{2}\bigg)}$

$\rightarrow\rm{cosA+cosB=2cos\bigg(\dfrac{A+B}{2}\bigg)cos\bigg(\dfrac{A-B}{2}\bigg)}$

$\rightarrow\rm{sin^{2}A+cos^{2}A=1}$

$\rightarrow\rm{cos(A+B)=cosAcosB-sinAsinB}$

$\rightarrow\rm{cos2A=\dfrac{1-tan^{2}A}{1+tan^{2}A}}$

$\rule{190}{1}$

On squaring given equations, we get

$\longrightarrow\rm{(cos\:\theta+cos\:\phi)^{2}=a^{2}}$

$\longrightarrow\rm{cos^{2}\theta+cos^{2}\phi+2cos\theta cos\phi=a^{2}}--(1)$

$\longrightarrow\rm{(sin\:\theta+sin\:\phi)^{2}=b^{2}}$

$\longrightarrow\rm{sin^{2}\theta+sin^{2}\phi+2sin\theta sin\phi=b^{2}}--(2)$

On adding (1) and (2), we get

$\sf{cos^{2}\theta+cos^{2}\phi+2cos\theta cos\phi+sin^{2}\theta+sin^{2}\phi+2sin\theta sin\phi=a^{2}+b^{2}}$

$\sf{sin^{2}\theta+cos^{2}\theta+sin^{2}\phi+cos^{2}\phi+2cos\theta cos\phi+2sin\theta sin\phi=a^{2}+b^{2}}$

$\sf{1+1+2(cos\theta cos\phi+sin\theta sin\phi)=a^{2}+b^{2}}$

$\sf{2+2(cos\theta cos\phi+sin\theta sin\phi)=a^{2}+b^{2}}$

$\sf{2(1+cos\theta cos\phi+sin\theta sin\phi)=a^{2}+b^{2}}$

$\sf{1+cos\theta cos\phi+sin\theta sin\phi=\dfrac{a^{2}+b^{2}}{2}}$

$\sf{cos\theta cos\phi+sin\theta sin\phi=\dfrac{a^{2}+b^{2}}{2}-1}$

$\sf\pink{cos\theta cos\phi+sin\theta sin\phi=\dfrac{a^{2}+b^{2}-2}{2}}--(3)$

$\rule{190}{1}$

Also, given equations can be written as

$\longrightarrow\rm{cos\:\theta+cos\:\phi=a}$

$\rightarrow\rm{2cos\bigg(\dfrac{\theta+\phi}{2}\bigg)cos\bigg(\dfrac{\theta-\phi}{2}\bigg)=a}--(4)$

$\longrightarrow\rm{sin\:\theta+sin\:\phi=b}$

$\rightarrow\rm{2sin\bigg(\dfrac{\theta+\phi}{2}\bigg)cos\bigg(\dfrac{\theta-\phi}{2}\bigg)=b}--(5)$

On dividing (5) by (4), we get

$\longrightarrow\rm{\dfrac{\cancel{2}sin\bigg(\dfrac{\theta+\phi}{2}\bigg)\cancel{cos\bigg(\dfrac{\theta-\phi}{2}\bigg)}}{\cancel{2}cos\bigg(\dfrac{\theta+\phi}{2}\bigg)\cancel{cos\bigg(\dfrac{\theta-\phi}{2}\bigg)}}=\dfrac{b}{a}}$

$\longrightarrow\rm{tan\bigg(\dfrac{\theta+\phi}{2}\bigg)=\dfrac{b}{a}}---(6)$

$\rule{190}{1}$

Now,

$\longrightarrow\rm{cos(\theta+\phi)=\dfrac{1-tan^{2}\bigg(\dfrac{\theta+\phi}{2}\bigg)}{1+tan^{2}\bigg(\dfrac{\theta+\phi}{2}\bigg)}}$

From (6), we get

$\longrightarrow\rm{cos(\theta+\phi)=\dfrac{1-\bigg(\dfrac{b}{a}\bigg)^{2}}{1+\bigg(\dfrac{b}{a}\bigg)^{2}}}$

$\longrightarrow\rm{cos(\theta+\phi)=\dfrac{\dfrac{a^{2}-b^{2}}{\cancel{a^{2}}}}{\dfrac{a^{2}+b^{2}}{\cancel{a^{2}}}}}$

$\longrightarrow\rm{cos(\theta+\phi)=\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}}$

$\longrightarrow\rm\pink{cos\theta cos\phi-sin\theta sin\phi=\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}}--(7)$

$\rule{190}{1}$

On adding (3) and (7), we get

$\sf{cos\theta cos\phi+sin\theta sin\phi+cos\theta cos\phi-sin\theta sin\phi=\dfrac{a^{2}+b^{2}-2}{2}+\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}}$

$\rm{2cos\theta cos\phi=\dfrac{a^{2}+b^{2}-2}{2}+\dfrac{a^{2}-b^{2}}{a^{2}+b^{2}}}$

$\sf{2cos\theta cos\phi=\dfrac{a^{2}+b^{2}(a^{2}+b^{2}-2)+2(a^{2}-b^{2})}{2(a^{2}+b^{2})}}$

$\sf{cos\theta cos\phi=\dfrac{(a^{2}+b^{2})^{2}-2(a^{2}+b^{2})+2(a^{2}-b^{2})}{4(a^{2}+b^{2})}}$

$\sf{cos\theta cos\phi=\dfrac{(a^{2}+b^{2})^{2}-2a^{2}-2b^{2}+2a^{2}-2b^{2}}{4(a^{2}+b^{2})}}$

$\longrightarrow\rm\green{cos\theta cos\phi=\dfrac{(a^{2}+b^{2})^{2}-4b^{2}}{4(a^{2}+b^{2})}}$

Hence, the correct option is (D).

Answer 2

Solution:-

cos∅+cosø=a......i)

sin∅+sinø=b.,......ii)

sqaure both side

and then add them

cos^2∅+sin^2∅+cos^2ø+sin^2ø+2sin∅sinø+2cos∅cosø=a^2+b^2

2+2(sin∅sinø+cos∅cosø)=a^2+b^2

2+2{(1-cos^2∅)^1/2(1-cos^2ø)^1/2+cos∅cosø}=a^2+b^2

(1-cos^2∅)^1/2(1-cos^2ø)^1/2=(a^2+b^2-2)/2-cos∅cos∅

sqaure on both side

(1-cos^2∅)(1-cos^2ø)={(a^2+b^2-2)/2-cosøcos∅}^2

=>1-cos^2ø-cos^2∅+cos^2øcos^2∅

={(a^2+b^2-2)/2-cosøcos∅}^2

put cos∅cosø=x

=>1-(a^2-2x)+x^2={(a^2+b^2-2)/2-x}^2

=>1-a^2+2x+x^2={(a^2+b^2-2)^2}/4+x^2

-(a^2+b^2-2)x

=>(2+a^2+b^2-2)x={(a^2+b^2-2)^2/4}

=>x=(a^2+b^2-4)^2/4(a^2+b^2)

as x=cos∅cosø

so

cosøcos∅=(a^2+b^2-2)^2/4(a^2+b^2)

cosøcos∅=(a^4+b^4+2a^2b^2-

4b^2-4a^2)/4(a^2+b^2)

cos∅cosø={(a^2+b^2)^2-4b^2}/4(a^2+b^2)

hence option d)