3. trignometrical ratios,functions and identities — ગણિત ધોરણ 11 સાયન્સ — Question

$ = {\cos ^2}(A - B) + {\cos ^2}B$

$ - \cos \,(A - B)\,\left\{ {\cos (A - B) + \cos (A + B)} \right\}$

$ = {\cos ^2}B - \cos \,(A - B)\,\,\cos \,\,(A + B)$

$ = {\cos ^2}B - ({\cos ^2}A - {\sin ^2}B) = 1 - {\cos ^2}A$

Hence it depends on $A.$

Trick : Put two different values of $A$.

Let $A = {90^o},$ then the value of expression will be ${\sin ^2}B + {\cos ^2}B = 1$

Now put $A = {0^o}$, then the value of expression will be ${\cos ^2}B + {\cos ^2}B - 2\,\,{\cos ^2}B = 0$

It means that the expression has different values for different $A$ 

$i.e.$ it depends on $A.$

Now similarly for $B = {90^o},$

the value of expression will be ${\sin ^2}A + 0 - 0$

$ = {\sin ^2}A$ અને at $B\,\, = {0^o}$

the value of expression will be ${\cos ^2}A + 1 - 2{\cos ^2}A = {\sin ^2}A$.

Hence, the expression has the same value for different values of $B$,

so it does not depend on $B.$