જો tan x = frac{{2b}}{{a - c}}(a ne c),

y | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b) We have,

$y + z = a({\cos ^2}x + {\sin ^2}x) + c({\sin ^2}x + {\cos ^2}x) = a + c$

$(	herefore {m{solution\,\, is\,\, (b)}} )$

$y - z

Vedclass · Accepted Answer

$y + z = a + c$

Vedclass · Answer

(b) We have,

$y + z = a({\cos ^2}x + {\sin ^2}x) + c({\sin ^2}x + {\cos ^2}x) = a + c$

$(	herefore {m{solution\,\, is\,\, (b)}} )$

$y - z = a({\cos ^2}x - {\sin ^2}x) + 4b\sin x\cos x$

$ - c({\cos ^2}x - {\sin ^2}x)$

$ = (a - c)\cos 2x + 2b\sin 2x$

$ = (a - c).\,\left( {\frac{{1 - {{	an }^2}x}}{{1 + {{	an }^2}x}}} ight) + 2b.\left( {\frac{{2	an x}}{{1 + {{	an }^2}x}}} ight)$

$ = (a - c).\left\{ {\frac{{1 - 4{b^2}/{{(a - c)}^2}}}{{1 + 4{b^2}/{{(a - c)}^2}}}} ight\} + 2b.\left\{ {\frac{{2.2b/(a - c)}}{{1 + 4{b^2}/{{(a - c)}^2}}}} ight\}$

Since $	an x = \frac{{2b}}{{(a - c)}}$,

$	herefore y - z = \frac{{(a - c).\{ {{(a - c)}^2} - 4{b^2}\} + 8{b^2}(a - c)}}{{{{(a - c)}^2} + 4{b^2}}}$

$ = \frac{{(a - c){{(a - c)}^2} + 4{b^2}}}{{\{ {{(a - c)}^2} + 4{b^2}\} }} = (a - c)$

$ \Rightarrow y 
e z\,,\,(a 
e c)$

જો $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$

$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$

અને $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ તો

Similar Questions

જો $A$ એ તૃતીય ચરણમાં હોય અને $3\ tanA - 4 = 0$ થાય તો $5\ sin\ 2A + 3\ sinA + 4\ cosA$ =

$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $

જો $\cos A = \frac{3}{4}$, તો $32\sin \frac{A}{2}\cos \frac{5}{2}A = $

જો $A, B, C $ એ ધન લઘુકોણ હોય તો $A + B + C = \pi $ અને $\cot A\,\cot \,B\,\cot \,C = K,$ તો

જો $\sin A = n\sin B,$ તો $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $

જો $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$ $y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$ અને $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ તો