Prove that: (i) cos^2 A + cos^2 B – 2 cos A cos B cos (A +B) = sin^2 (A + B)

(ii)

RD Sharma Solutions for Class 11 Maths Chapter 7 – Values of Trigonometric Functions at Sum or Difference of Angles image- 13

Solution:

(i)

$co{{s}^{2}}A+co{{s}^{2}}B - 2cosAcosBcos\left(A+B \right)=si{{n}^{2}}\left(A+B\right)$

Let us consider LHS:

$co{{s}^{2}}A\text{ }+\text{ }co{{s}^{2}}B\text{}-\text{}2\text{}cos\text{}A\text{}cos\text{}B\text{}cos\text{}\left(A\text{}+B\right)$

$co{{s}^{2}}A\text{}+\text{}1\text{}-\text{}si{{n}^{2}}B\text{}-\text{}2\text{}cos\text{}A\text{}cos\text{}B\text{}cos\text{}\left(A\text{}+B \right)$

$1\text{}+co{{s}^{2}}A\text{}\text{}si{{n}^{2}}B\text{}-\text{}2\text}cos\text{}A\text{}cos\text{}B\text{}cos\text{}\left(A\text{}+B\right)$

Since,

$co{{s}^{2}}A\text{ }-\text{ }si{{n}^{2}}B\text{}=-\text{}cos\text{}\left(A\text{}+B\right)\text{}cos\text{ }\left( A\text{ }B \right)$

Therefore,

$1+~cos\left( A+B \right)cos\left( A-B \right)-2cosAcosBcos\left( A+B \right)$

$1\text{ }+\text{ }cos\text{ }\left( A\text{ }+B \right)\text{ }\left[ cos\text{ }\left( A\text{ }-B \right)\text{ }-\text{ }2\text{ }cos\text{ }A\text{ }cos\text{ }B \right]$

Since,

$cos\text{ }\left( A\text{ }-\text{ }B \right)\text{ }=\text{ }cos\text{ }A\text{ }cos\text{ }B\text{ }+\text{ }sin\text{ }A\text{ }sin\text{ }B.$

Therefore,

$1+cos\left( A+B \right)\text{ }\left[ cosA\text{ }cosB+sinA\text{ }sinB\text{ }2\text{ }cosAcosB \right]$

$1\text{ }+\text{ }cos\text{ }\left( A\text{ }+B \right)\text{ }\left[ -cos\text{ }A\text{ }cos\text{ }B\text{ }+\text{ }sin\text{ }A\text{ }sin\text{ }B \right]$