Download App

Transformation Formulae — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTransformation Formulae5 Marks
Question
Prove that:
$\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}=\cot6\text{A}$
✓

Answer

We have,
$\text{LHS}=\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}$
$=\ \frac{-(\sin7\text{A}-\sin5\text{A})+(\sin8\text{A}-\sin4\text{A})}{-(\cos7\text{A}-\cos5\text{A})-(\cos8\text{A}-\cos4\text{A})}$
$=\ \frac{-\Big[2\sin\Big(\frac{7\text{A}-5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{8\text{A}-4\text{A}}{2}\Big)\cos\Big(\frac{8\text{A}+4\text{A}}{2}\Big)\Big]}{-2\sin\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\sin\Big(\frac{7\text{A}-5\text{A}}{2}\Big)-\Big[-2\sin\Big(\frac{8\text{A}+4\text{A}}{2}\Big)\sin\Big(\frac{8\text{A}-4\text{A}}{2}\Big)\Big]}$
$=\ \frac{-2\sin\text{A}\cos6\text{A}+2\sin2\text{A}\cos6\text{A}}{-2\sin6\text{A}\sin\text{A}+2\sin6\text{A}\sin2\text{A}}$
$=\ \frac{2\cos6\text{A}[-\sin\text{A}+\sin2\text{A}]}{-2\sin6\text{A}[-\sin\text{A}+\sin2\text{A}]}$
$=\ \frac{\cos6\text{A}}{\sin6\text{A}}$
$=\ \cot6\text{A}$
$=\ \text{RHS}$
$\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}=\cot6\text{A}$ Hence proved.

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 "5 Marks Questions" from Transformation Formulae→Transformation Formulae — all question types→MATHS STD 11 Science question papers→Rajasthan Board STD 11 Science question papers→Rajasthan Board question paper generator→

Similar questions

Show that $\frac { 1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + \ldots \ldots + n \times ( n + 1 ) ^ { 2 } } { 1 ^ { 2 } \times 2 + 2 ^ { 2 } \times 3 + \ldots \ldots + n ^ { 2 } ( n + 1 ) } = \frac { 3 n + 5 } { 3 n + 1 }$
→
If a, b, c, are in G.P., prove that the following are also in G.P.
$\text{a}^2+\text{b}^2,\text{ab}+\text{bc},\text{b}^2+\text{c}^2$
→
For three sets A, B and C, show that.
$\text{A}\subset\text{B}\Rightarrow\text{C}-\text{B}\subset\text{C}-\text{A}.$
→
For any two sets A and B, prove that
$(\text{A}\cap\text{B})-\text{B = A}- \text{B}$
→
Evaluate the following limit:
$\lim\limits_{\text{h}\rightarrow0}\frac{\sqrt{\text{x}+\text{h}}-\sqrt{\text{x}}}{\text{h}},\text{x}\ne0$
→
Prove that:
$\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$
→
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{8}}\frac{\cot4\text{x}-\cos4\text{x}}{(\pi-8\text{x})^3}$
→
If $2\tan\alpha=3\tan\beta,$ prove that $\tan(\alpha-\beta)=\frac{\sin2\beta}{5-\cos2\beta}$
→
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow{\frac{\pi}{4}}}\frac{\sqrt{2}-\cos\text{x}-\sin\text{x}}{\big(\frac{\pi}{4}-\text{x}\big)^2}$
→
The perpendicular from the origin to the line y = mx + c meets it at the point (-1, 2). Find the values of m and c.
→