No products in the cart.

2 sin 68/cos 22 -2 cot …

CBSE, JEE, NEET, CUET

CBSE, JEE, NEET, CUET

Question Bank, Mock Tests, Exam Papers

NCERT Solutions, Sample Papers, Notes, Videos

2 sin 68/cos 22 -2 cot 15/5tan 75 =5tan45 tan 20 tan 40 tan 5 tan 7/5
  • 1 answers

Sia ? 6 years, 5 months ago

{tex}\frac{{2\sin 68^\circ }}{{\cos 22^\circ }} - \frac{{2\cot 15^\circ }}{{5\tan 75^\circ }} - \frac{{3\tan 45^\circ \tan 20^\circ \tan 40^\circ \tan 50^\circ \tan 70^\circ }}{5}{/tex}
{tex} = \frac{{2\sin \left( {90^\circ - 22^\circ } \right)}}{{\cos 22^\circ }} - \frac{{2\cot \left( {90^\circ - 75^\circ } \right)}}{{5\tan 75^\circ }}{/tex} {tex} - \frac{{3\tan 45^\circ \tan \left( {90^\circ - 70^\circ } \right)\tan \left( {90^\circ - 50^\circ } \right)\tan 50^\circ \tan 70^\circ }}{5}{/tex}
{tex} = \frac{{2\cos 22^\circ }}{{\cos 22^\circ }} - \frac{{2\tan 75^\circ }}{{5\tan 75^\circ }} - \frac{{3 \times 1 \times \cot 70^\circ \times \tan 50^\circ \times \frac{1}{{\cot 50^\circ }} \times \frac{1}{{\cot 70^\circ }}}}{5}{/tex}
{tex} = 2 - \frac{2}{5} - \frac{3}{5}{/tex}
{tex} = \frac{{10 - 2 - 3}}{5} = \frac{5}{5} = 1{/tex}

https://examin8.com Test

Related Questions

Find the nature of quadratic equation x^2 +x -5 =0
  • 0 answers
(A + B )²
  • 1 answers
Venu Gopal has twice
  • 0 answers
sin60° cos 30°+ cos60° sin 30°
  • 2 answers
X-y=5
  • 1 answers
Prove that root 8 is an irration number
  • 2 answers

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App