tan⁡81∘−tan⁡63∘−tan⁡27∘+tan⁡9∘ equals

# $\mathrm{tan}{81}^{\circ }-\mathrm{tan}{63}^{\circ }-\mathrm{tan}{27}^{\circ }+\mathrm{tan}{9}^{\circ }$ equals

1. A

6

2. B

0

3. C

2

4. D

4

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$\begin{array}{l}\mathrm{tan}{81}^{\circ }-\mathrm{tan}{63}^{\circ }-\mathrm{tan}{27}^{\circ }+\mathrm{tan}{9}^{\circ }=\left[\mathrm{tan}{81}^{\circ }+\mathrm{tan}{9}^{\circ }\right)-\left[\mathrm{tan}\left({63}^{\circ }\right)+\mathrm{tan}\left({27}^{\circ }\right)\right]\\ =\left[\frac{\mathrm{sin}{81}^{\circ }}{\mathrm{cos}{81}^{\circ }}+\frac{\mathrm{sin}{9}^{\circ }}{\mathrm{cos}{9}^{\circ }}\right]-\left[\frac{\mathrm{sin}{63}^{\circ }}{\mathrm{cos}{63}^{\circ }}+\frac{\mathrm{sin}{27}^{\circ }}{\mathrm{cos}{27}^{\circ }}\right]\\ =\left[\frac{\mathrm{sin}{81}^{\circ }\mathrm{cos}{9}^{\circ }+\mathrm{sin}{9}^{\circ }\mathrm{cos}{81}^{\circ }}{\mathrm{cos}{9}^{\circ }\mathrm{cos}{81}^{\circ }}\right]-\left[\frac{\mathrm{sin}{63}^{\circ }\mathrm{cos}{27}^{\circ }+\mathrm{sin}{27}^{\circ }\mathrm{cos}{63}^{\circ }}{\mathrm{cos}{27}^{\circ }\mathrm{cos}{63}^{\circ }}\right]\\ =\frac{\left[\mathrm{sin}\left({81}^{\circ }+{9}^{\circ }\right)\right]}{\mathrm{cos}{9}^{\circ }\mathrm{cos}\left({90}^{\circ }-{9}^{\circ }\right)}-\frac{\left[\mathrm{sin}\left({63}^{\circ }+{27}^{\circ }\right)\right]}{\mathrm{cos}{27}^{\circ }\mathrm{cos}\left({90}^{\circ }-{27}^{\circ }\right)}\\ =\frac{\mathrm{sin}{90}^{\circ }}{\mathrm{cos}{9}^{\circ }\mathrm{sin}{9}^{\circ }}-\frac{\mathrm{sin}{90}^{\circ }}{\mathrm{cos}{27}^{\circ }\mathrm{sin}{27}^{\circ }}\\ =\frac{1}{\mathrm{sin}{9}^{\circ }\mathrm{cos}{9}^{\circ }}×\frac{2}{2}-\frac{1}{\mathrm{sin}{27}^{\circ }\mathrm{cos}{27}^{\circ }}×\frac{2}{2}\\ =\frac{2}{\mathrm{sin}{18}^{\circ }}-\frac{2}{\mathrm{sin}{54}^{\circ }}\\ =2\left[\frac{\mathrm{sin}{54}^{\circ }-\mathrm{sin}{18}^{\circ }}{\mathrm{sin}{18}^{\circ }\mathrm{sin}{54}^{\circ }}\right]=2\left[\frac{2\mathrm{cos}{36}^{\circ }\mathrm{sin}{18}^{\circ }}{\mathrm{sin}{18}^{\circ }\mathrm{cos}{36}^{\circ }}\right]=4\end{array}$

## Related content

 How to Score 100 in Class 6 Maths using NCERT Solutions TS EAMCET Previous Year Question Papers CBSE Class 8 English Syllabus Academic Year 2023-2024 CBSE Class 7 English Syllabus Academic Year 2023-2024 CBSE Worksheets for Class 7 with Answers COMEDK UGET Mock Test 2024 (Available) – Free Mock Test Series Indian tribes Maurya Empire CBSE Class 10 Science Important Topics – You Should Not Miss in Board Exam 2024 CBSE Class 6 Social Science: Important Tips and Topics

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)