Download App

Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles1 Mark
MCQ
If $\text{A+B+C}=\pi,$ then $\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})$ is equal to:
  • A
    0
  • ✓
    -1
  • C
    1
  • D
    None of these
✓

Answer

Correct option: B.
-1
$\pi=180^\circ$
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\pi-(\text{A+B}))-\sin\text{B}\sin(\pi(\text{A+B}))}{\cos\text{A}}$
We know that, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta,$
$\therefore\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\text{A+B})-\sin\text{B}\sin(\text{A+B})}{\cos\text{A}}$
Now, using the identities $\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ and $\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B},$ we get
$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}\cos\text{B}^2+\cos\text{B}\sin\text{A}\sin\text{B}-\sin\text{B}\sin\text{A}\cos\text{B}-\sin^2\cos\text{A}}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}(\cos^2\text{B}+\sin^2\text{B})}{\cos\text{A}}$
$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}}{\cos\text{A}}=-1$

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 Trigonometric Ratios Of Multiple And Sub Multiple Angles→Trigonometric Ratios Of Multiple And Sub Multiple Angles — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

If $\frac{\text{T}_{2}}{\text{T}_{3}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}}$ and $\frac{\text{T}_{3}}{\text{T}_{4}}$ in the expansion of $(\text{a}+\text{b})^{\text{n}+3}$ are equal, then $n =$
→
Let $\mathrm{A}$ be the set of all points $(\alpha, \beta)$ such that the area of triangle formed by the points $(5,6),(3,2)$ and $(\alpha, \beta)$ is $12\, square\, units.$ Then the least possible length of a line segment joining the origin to a point in $A,$ is :
→
Locus of foot of normal drawn from any focus to variable tangent of hyperbola $4x^2-9y^2-8x- 18y = 41$ will be
→
In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament?
→
The sum of the first $20$ terms common between the series $3 +7 + 1 1 + 15+ ... ......$ and $1 +6+ 11 + 16+ ......$, is
→
The sum of all $4$ digit numbers that can be formed by using the digits $2, \,4,\, 6,\, 8$ (repetition of digits not allowed) is
→
Eccentricity of conjugate hyperbola of $16x^2 - 9y^2 - 32x - 36y - 164 = 0$ will be-
→
If ${\left( {10} \right)^9} + 2{\left( {11} \right)^1}{\left( {10} \right)^8} + 3{\left( {11} \right)^2}{\left( {10} \right)^7} + ..\;.\;.\;.\; + 10\left( {{{11}^9}} \right) = \;k{\left( {10} \right)^9}$ ,then $k $ is equal to 
→
The number of $4$ letter words (with or without meaning) that can be formed from the eleven letters of the word $'EXAMINATION'$ is
→
The product of the lengths of perpendiculars drawn from any point on the hyperbola ${x^2} - 2{y^2} - 2 = 0$ to its asymptotes is
→