Prove that

Asked by debabratap261 | 18th Jan, 2019, 10:24: AM

Expert Answer:

begin mathsize 18px style fraction numerator secθ plus tanθ plus 1 over denominator secθ minus tanθ plus 1 end fraction
equals fraction numerator begin display style secθ plus tanθ plus sec squared straight theta minus tan squared straight theta end style over denominator begin display style secθ minus tanθ plus 1 end style end fraction
equals fraction numerator begin display style secθ plus tanθ plus open parentheses secθ minus tanθ close parentheses open parentheses secθ plus tanθ close parentheses end style over denominator begin display style secθ minus tanθ plus 1 end style end fraction
equals fraction numerator begin display style open parentheses secθ plus tanθ close parentheses open parentheses 1 plus secθ minus tanθ close parentheses end style over denominator begin display style secθ minus tanθ plus 1 end style end fraction
equals secθ plus tanθ
equals fraction numerator 1 plus sinθ over denominator cos begin display style straight theta end style end fraction
equals fraction numerator begin display style 1 plus sinθ open parentheses 1 minus sinθ close parentheses end style over denominator begin display style cos straight theta open parentheses 1 minus sinθ close parentheses end style end fraction
equals fraction numerator begin display style open parentheses 1 minus sin squared straight theta close parentheses end style over denominator begin display style cosθ open parentheses 1 minus sinθ close parentheses end style end fraction
equals fraction numerator begin display style open parentheses cos squared straight theta close parentheses end style over denominator begin display style cosθ open parentheses 1 minus sinθ close parentheses end style end fraction
equals fraction numerator cosθ over denominator 1 minus sinθ end fraction end style

Answered by Arun | 18th Jan, 2019, 12:00: PM