Home/Class 10/Maths/

Question and Answer

If a and b are real number such that \( acos\theta +bsin\theta =4\) and \( asin\theta -bcos\theta =3\), then \( {(a}^{2}+{b}^{2})\) is
(A) 7
(B) 12
(C) 25
(D) \( \sqrt{12}\)

Answer

(C)
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Answer, Download the App
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play

Solution

\( acos\theta +bsin\theta =4 \) ______ (1)
\( asin\theta -bcos\theta =3\) _________(2)
\( (1{)}^{2}+(2{)}^{2}\)
\( =(acos\theta +bsin\theta {)}^{2}(asin\theta -bcos\theta {)}^{2}={4}^{2}+{3}^{2}  \)
\( {a}^{2}co{s}^{2}\theta +{b}^{2}si{n}^{2}\theta +ab\;sin\theta\;cos\theta +{a}^{2}si{n}^{2}\theta +{b}^{2}co{s}^{2}\theta -2ab\;sin\theta\;cos\theta =16+9 \)
\( {a}^{2}(co{s}^{2}\theta +si{n}^{2}\theta )+{b}^{2}(si{n}^{2}\theta +co{s}^{2}\theta )=25\)
\( {a}^{2}+{b}^{2}=25\)
\( (\therefore\;si{n}^{2}\theta +co{s}^{2}\theta =1\)
To Keep Reading This Solution, Download the APP
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
To Keep Reading This Solution, Download the APP
4.6
star pngstar pngstar pngstar pngstar png
Review from Google Play
Correct38
Incorrect0
Watch More Related Solutions
The value of \( sin\theta .cos\theta (tan\theta +cot\theta )\) is
(A) 1
(B) 2
(C) -1
(D) 0
If \( sin\theta =\frac{3}{5},\) then the value of \( (tan\theta +sec\theta {)}^{2} \)is
(A) 2
(B) 4
(C) 6
(D) 8
\( tan35°. tan40°.tan45°.tan50°tan55° \) is equal to
(A) 1
(B) 2
(C) 0
(D) 3
If \( tan\theta =\frac{3}{4},\) evaluate \( \frac{4sin\theta -2cos\theta }{4sin\theta +3cos\theta }\) is
(A) \( 1∕3\)
(B) \( 1∕4\)
(C) \( 1∕5\)
(D) \( 1∕6\)
If A is an acute angle and \( tanA=\frac{5}{12}, \)the value of case A is
(A) \( \frac{11}{5}\)
(B) \( \frac{13}{5}\)
(C) \( \frac{16}{5}\)
(D) \( \frac{17}{5}\)
If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( 4{x}^{2}+2x-1=0\) then the value of \( \sum _{r=1}^{\infty }\left({\alpha }^{r}+{\beta }^{r}\right)\) is:
(a) \( 2\)
(b) \( 3\)
(c) \( 6\)
(d) \( 0\)
\( \frac{cos\left(90°-\theta \right)cos\theta }{tan\theta }+\frac{co{s}^{2}(90°-\theta )}{1}=\)
(A) 0
(B) 1
(C) 2
(D) 3
\( \frac{sin\theta .cos\theta\;cos(90°-\theta )}{sin(90°-\theta )}+\frac{sin\theta\;cos\theta .sin(90°-\theta )}{cos(90°-\theta )}\) is equal to
(A) 0
(B) 1
(C) 2
(D) 3
\( \frac{sin\theta }{sin(90°-0)}+\frac{cos \theta }{cos(90°-0)}\) is equal to
(A) \( sin\theta .cos\theta \)
(B) \( sec\theta .cosec \theta \)
(C) 1
(D) 2
\( \frac{cos35°}{sin55°}+\frac{tan27°tan63°}{sin30°}-3ta{n}^{2}60°\) is equal to
(A) 6
(B) 6
(C) -6
(D) 3

Load More