alpha+beta=pi/2 এবং beta+gamma=alpha হলে tanalpha-এর মান হবে-
正切函数的诱导公式(1)$$ \tan ( 2 \pi + \alpha ) = \tan \alpha $$ ;(2)$$ \tan ( - \alpha ) = - \tan \alpha $$(3)$$ \tan ( 2 \pi - \alpha ) = - \tan \alpha $$(4)$$ \tan ( \pi - \alpha ) = - \tan \alpha $$;(5)$$ \tan ( \pi + \alpha...
【解析】 答案: $$ ( 1 ) \tan \alpha ; ( 2 ) - \tan \alpha ; ( 3 ) - \tan \alpha ; ( 4 ) - \tan \alpha ; ( 5 ) \tan \alpha $$ 分析: (1)$$ \tan ( 2 \pi + \alpha ) = \tan \alpha $$ (2)$$ \tan ( - \alpha ) = - \tan \alpha , $$ (3)$$...
1/(secalpha-tanalpha)-1/cosalpha=1/cosalpha-1/(secalpha+tanalpha) (sinα+cosα)(tanα+cotα)=secα+cosecα View Solution Ifsecα+tanα=m,thensec4α−tan4α−2secαtanαis ___. मान ज्ञात कीजिए -∣∣ ∣∣01secαtanα−secαtanα101∣...
【解析】 答案: $$ 1 ) \tan \alpha ; ( 2 ) - \tan \alpha ; ( 3 ) - \tan \alpha ; ( 4 ) - \tan \alpha ; ( 5 ) \tan \alpha $$ 分析: (1)$$ \tan ( 2 \pi + \alpha ) = \tan \alpha , $$ (2)$$ \tan ( - \alpha ) = - \tan \alpha , $$ (3)$$...
Step by step video & image solution for যদি tan alpha = x/y [ 0 lt alpha lt pi/2] হয়, তবে দেখাও যে, x cosec ""(alpha)/3 - y sec ""(alpha)/3 = 2 sqrt (x^2 + y^2) by Maths experts to help you in doubts & scoring excellen...
解析:cosalpha;=-35,alpha;isin;(pi;2,pi;),所以sinalpha;=45,there4;tanalpha;=sinalpha;cosalpha;=-43.相关知识点: 试题来源: 解析 答案:-43 解析:cosalpha;=-35,alpha;isin;(pi;2,pi;),所以sinalpha;=45,there4;tanalpha;=sinalpha;cosalpha;=-43.反馈...
正切函数的诱导公式(1)$$ \tan ( 2 \pi + \alpha ) = \_ ; $$(2)$$ \tan ( - \alpha ) = \_ ; $$(3)$$ \tan ( 2 \pi - \alpha ) = \_ ; $$⑷)$$ \tan ( \pi - \alpha ) = \_ ; $$(5)$$ \tan ( \pi + \alpha ) = \_ ; $$ 相关知识点: ...
If alpha=(2 pi)/(7), then tan alpha tan2 alpha+tan2 alpha tan4 alpha+tan4 alpha tan alpha View Solution [ 7.If 0ltalphaltbetalt(pi)/(2) then [ 1) (tan beta)/(tan alpha)lt(alpha)/(beta), 2) (tan beta)/(tan alpha)gt(alpha)/(beta) 3) (tan alpha)/(tan beta)lt(alp...
14.【解析】由$$ \tan ( \pi + 2 \alpha ) = - \frac { 4 } { 3 } $$得$$ \tan 2 \alpha = - \frac { 4 } { 3 } $$, 又$$ \tan 2 \alpha = \frac { 2 \tan \alpha } { 1 - \tan ^ { 2 } \alpha } = - \frac { 4 } { 3 } , $$ 解得$$ \tan...