Download App

STD 11 - 3. trignometrical ratios,functions and identities — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 3. trignometrical ratios,functions and identities1 Mark
MCQ
$1 + \cos \,{56^o} + \cos \,{58^o} - \cos {66^o} = $
  • A
    $2\,\cos {28^o}\,\cos \,{29^o}\,\cos \,{33^o}$
  • B
    $4\,\cos {28^o}\,\cos \,{29^o}\,\cos \,{33^o}$
  • ✓
    $4\,\cos {28^o}\,\cos \,{29^o}\,\sin {33^o}$
  • D
    $2\,\cos {28^o}\,\cos \,{29^o}\,\sin \,{33^o}$
✓

Answer

Correct option: C.
$4\,\cos {28^o}\,\cos \,{29^o}\,\sin {33^o}$
c
(c) $1 + \cos 56^\circ + \cos 58^\circ - \cos 66^\circ $

$ = 2{\cos ^2}28^\circ + 2\sin 62^\circ .\sin 4^\circ $

$ = 2{\cos ^2}28^\circ + 2\cos 28^\circ .\sin 4^\circ $

$ = 2\cos 28^\circ (\cos 28^\circ + \cos 86^\circ )$ 

$ = 2\cos 28^\circ .2\cos 57^\circ \cos 29^\circ $

$ = 4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.

Aliter : Apply the conditional identity

$\cos A + \cos B - \cos C = - 1 + 4\cos \frac{A}{2}\cos \frac{B}{2}\sin \frac{C}{2}$                                                              $[\, \because 56^\circ  + 58^\circ  + 66^\circ  = 180^\circ ]$

We get the value of required expression equal to $4\cos 28^\circ \cos 29^\circ \sin 33^\circ $.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 11 - 3. trignometrical ratios,functions and identities→STD 11 - 3. trignometrical ratios,functions and identities — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Find the probability that the two digit number formed by digits $1, 2, 3, 4, 5$ is divisible by $4$ (while repetition of digit is allowed)
→
$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} |(1 - \sin x)\tan x$ is
→
In an isosceles triangle $ABC$, the vertex $A$ is $(6,1)$ and the equation of the base $BC$ is $2 x + y =4$. Let the point $B$ lie on the line $x+3 y=7$. If $(\alpha, \beta)$ is the centroid $\triangle A B C$, then $15(\alpha+\beta)$ is equal to
→
The locus of centre of a circle passing through $(a, b)$ and cuts orthogonally to circle ${x^2} + {y^2} = {p^2}$, is
→
Let $n(A) = m,$ and $n(B) = n,$ Then the total number of non$-$empty relations that can be defined from $A$ to $B$ is:
→
If $A+B+C=90^{\circ}$, then the value of $\tan (B+C)$ is :
→
If $\cos A = \frac{{\sqrt 3 }}{2},$ then $\tan 3A = $
→
Circle$(s)$ touching $x$-axis at a distance $3$ from the origin and having an intercept of length $2 \sqrt{7}$ on $y$-axis is (are)

$(A)$ $x^2+y^2-6 x+8 y+9=0$

$(B)$ $x^2+y^2-6 x+7 y+9=0$

$(C)$ $x^2+y^2-6 x-8 y+9=0$

$(D)$ $x^2+y^2-6 x-7 y+9=0$

→
If $4\text{x} + 3<6\text{x} + 7$, then$\text{ x}\in$
→
Let $f (x) = \left[ {\begin{array}{*{20}{c}}  {\tfrac{{{{\tan }^2}\,\,\{ \,x\,\} }}{{{x^2}\,\, - \,\,{{[\,x\,]}^2}}}\,\,\,\,} \\   {1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\   {\sqrt {\{ \,x\,\} \,\,\,\cot \,\,\{ \,x\,\} } } \end{array}} \right.\,\,\,\,\,\begin{array}{*{20}{c}}  {for\,\,\,\,\,x\,\, > \,\,0} \\    {for\,\,\,\,\,x\,\, = \,\,0} \\   {for\,\,\,\,\,x\,\, < \,\,0} \end{array}$ where $[ x ]$ is the step up function and $\{ x \}$ is the fractional part function of $x$ , then :
→