Advertisements
Advertisements
Question
Advertisements
Solution
\[Here, \]
\[\tan\left( 82 . 5 \right)° = \tan\left( 90 - 7 . 5 \right)°\]
\[ = \cot\left( 7 . 5 \right)°\]
\[ = \frac{1}{\tan\left( 7 . 5 \right)°}\]
\[\text{ We know } , \]
\[\tan\left( \frac{x}{2} \right) = \frac{\text{ sin } x}{1 + \text{ cos } x}\]
\[\text{ On putting } x = 15° , \text{ we get } \]
\[\tan \left( \frac{15}{2} \right)^°= \frac{\sin15°}{1 + \cos15°}\]
\[ = \frac{\sin\left( 45 - 30 \right)° }{1 + \cos\left( 45 - 30 \right)°}\]
\[ = \frac{\sin45° \cos30° - \sin30° \cos45° }{1 + \cos45° \cos30° + \sin45° \sin30° }\]
\[ = \frac{\left( \frac{1}{\sqrt{2}} \right) \times \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \times \left( \frac{1}{\sqrt{2}} \right)}{1 + \left( \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} \right) + \left( \frac{1}{\sqrt{2}} \times \frac{1}{2} \right)}\]
\[ = \frac{\frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}}{1 + \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}}\]
\[ = \frac{\sqrt{3} - 1}{2\sqrt{2} + \sqrt{3} + 1}\]
\[\text{ Now } , \]
\[\tan\left( 82 . 5 \right)° = \frac{1}{\tan\left( 7 . 5 \right)° }\]
\[ = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1}\]
\[ = \frac{2\sqrt{2} + \sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}\]
\[ = \frac{\sqrt{3} + 1\left( 2\sqrt{2} + \sqrt{3} + 1 \right)}{\left( \sqrt{3} \right)^2 - 1^2}\]
\[ = \frac{2\sqrt{6} + 3 + \sqrt{3} + 2\sqrt{2} + \sqrt{3} + 1}{3 - 1}\]
\[ = \frac{2\sqrt{6} + 2\sqrt{3} + 2\sqrt{2} + 4}{2}\]
\[ = \sqrt{6} + \sqrt{3} + \sqrt{2} + 2\]
\[ = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6} . . . \left( 1 \right)\]
\[ = \sqrt{6} + \sqrt{3} + 2 + \sqrt{2}\]
\[ = \sqrt{3}\left( \sqrt{2} + 1 \right) + \sqrt{2}\left( \sqrt{2} + 1 \right)\]
\[ = \left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{2} + 1 \right) . . . \left( 2 \right)\]
\[\text{ From eqs } . \left( 1 \right) \text{ and } \left( 2 \right), \text{ we get} \]
\[ \tan\left( 82 . 5 \right)° = \left( \sqrt{3} + \sqrt{2} \right)\left( \sqrt{2} + 1 \right) = \sqrt{2} + \sqrt{3} + \sqrt{4} + \sqrt{6}\]
APPEARS IN
RELATED QUESTIONS
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
If \[\cos x = - \frac{3}{5}\] and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\sin \alpha = \frac{4}{5} \text{ and } \cos \beta = \frac{5}{13}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \frac{8}{\sqrt{65}}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
Prove that: \[\sin 5x = 5 \sin x - 20 \sin^3 x + 16 \sin^5 x\]
Prove that `tan x + tan (π/3 + x) - tan(π/3 - x) = 3tan 3x`
\[\tan x + \tan\left( \frac{\pi}{3} + x \right) - \tan\left( \frac{\pi}{3} - x \right) = 3 \tan 3x\]
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
Write the value of \[\cos\frac{\pi}{7} \cos\frac{2\pi}{7} \cos\frac{4\pi}{7} .\]
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\] is
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
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 \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\cos \left( 36° - A \right) \cos \left( 36° + A \right) + \cos \left( 54° - A \right) \cos \left( 54° + A \right)\] is
The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] 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 A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
