Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Here, \]
\[\tan\left( 82 . 5 \right)° = \tan\left( 90 - 7 . 5 \right)°\]
\[ = \cot\left( 7 . 5 \right)°\]
\[ = \frac{1}{\tan\left( 7 . 5 \right)°}\]
\[\text{ We know } , \]
\[\tan\left( \frac{x}{2} \right) = \frac{\text{ sin } x}{1 + \text{ cos } x}\]
\[\text{ On putting } x = 15° , \text{ we get } \]
\[\tan \left( \frac{15}{2} \right)^°= \frac{\sin15°}{1 + \cos15°}\]
\[ = \frac{\sin\left( 45 - 30 \right)° }{1 + \cos\left( 45 - 30 \right)°}\]
\[ = \frac{\sin45° \cos30° - \sin30° \cos45° }{1 + \cos45° \cos30° + \sin45° \sin30° }\]
\[ = \frac{\left( \frac{1}{\sqrt{2}} \right) \times \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \times \left( \frac{1}{\sqrt{2}} \right)}{1 + \left( \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{1}{2} \right)}\]
\[ = \frac{\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}}{1 + \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}}\]
\[ = \frac{\sqrt{3} - 1}{2\sqrt{2} + \sqrt{3} + 1}\]
\[\text{ Now } , \]
\[\tan\left( 82 . 5 \right)° = \frac{1}{\tan\left( 7 . 5 \right)° }\]
\[ = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}\]
\[ = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\]
\[ = \frac{\sqrt{3} + 1\left( 2\sqrt{2} + \sqrt{3} + 1 \right)}{\left( \sqrt{3} \right)^2 - 1^2}\]
\[ = \frac{2\sqrt{6} + 3 + \sqrt{3} + 2\sqrt{2} + \sqrt{3} + 1}{3 - 1}\]
\[ = \frac{2\sqrt{6} + 2\sqrt{3} + 2\sqrt{2} + 4}{2}\]
\[ = \sqrt{6} + \sqrt{3} + \sqrt{2} + 2\]
\[ = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} . . . \left( 1 \right)\]
\[ = \sqrt{6} + \sqrt{3} + 2 + \sqrt{2}\]
\[ = \sqrt{3}\left( \sqrt{2} + 1 \right) + \sqrt{2}\left( \sqrt{2} + 1 \right)\]
\[ = \left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{2} + 1 \right) . . . \left( 2 \right)\]
\[\text{ From eqs } . \left( 1 \right) \text{ and } \left( 2 \right), \text{ we get} \]
\[ \tan\left( 82 . 5 \right)° = \left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Prove that: \[\frac{\cos 2 x}{1 + \sin 2 x} = \tan \left( \frac{\pi}{4} - x \right)\]
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\cos^3 2x + 3 \cos 2x = 4\left( \cos^6 x - \sin^6 x \right)\]
Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]
If \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan \frac{x}{2}\] .
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(i)\[\sin \left( \alpha + \beta \right) = \frac{2ab}{a^2 + b^2}\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
If \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that: \[4 \left( \cos^3 10 °+ \sin^3 20° \right) = 3 \left( \cos 10°+ \sin 2° \right)\]
Prove that: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
If \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
If \[\text{ sin } x + \text{ cos } x = a\], then find the value of
If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\] and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
If α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is
The greatest value of sin x cos x is ______.
If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.
`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta = (2tantheta)/(1 + tan^2theta)]`.
If tanθ = `1/2` and tanΦ = `1/3`, then the value of θ + Φ is ______.
