tan(alpha+beta)=2 tan(alpha-beta)=1 tan2alpha=tan[(alpha+beta)+(alpha-beta)] =(tan(alpha+beta)+tan(alpha-beta))/(1-tan(alpha+beta).tan(alpha-beta)) =(2+1)/(1-2.1)=(3)/(-1)=-3
The expression (tan alpha+tan beta)cot(alpha+beta)+(tan alpha-tan beta)cot(alpha-beta) is independent of both alpha and beta. View Solution Prove that (tan alpha+tan beta)/(cot alpha+cot beta)=tan alpha tan beta View Solution Iftanα+tanβcotα+cotβ+{cos(α−β)+1}−1=1,t...
\beta } k \pi + \frac { \pi } { 2 } , k \in Z $$ [思考探究] 1.提示 不一定有意义.如$$ \alpha = \beta = \frac { \pi } { 4 } $$,此时tana,tanβ均有意义, 但$$ \alpha + \beta = \frac { \pi } { 2 } $$,此时函数$$ \tan ( \alpha + \beta ) $$无...
已知alpha ,beta 为锐角,且tanα 相关知识点: 试题来源: 解析 B tan 2alpha =tan [(alpha +beta )+(alpha -beta )], =frac(tan (alpha +beta )-tan (alpha -beta ))(1+tan (alpha +beta )tan (alpha -beta )), 又tan 2alpha =4tan (alpha -beta ), 设tan 2alpha =x,则tan (alpha -...
1. $$ \frac { \tan \alpha + \tan \beta } { 1 - \tan \alpha \tan \beta } $$2. $$ \frac { \tan \alpha - \tan \beta } { 1 + \tan \alpha \tan \beta } $$ 思考 解:$$ \frac { \tan \alpha + \tan \beta } { 1 - \tan \alpha \tan \beta } = \ta...
Similar Questions 1−cos2θ= View Solution প্রমাণ করো:tan(pi/4+theta)+tan(pi/4-theta)=2sec2theta 03:47 Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class...
যদি tan(alpha+beta)=a+b এবং tan(alpha-beta)=a-b হয় তবে দেখাও যে atanalpha-btanbeta=a^2-b^2
1. 两角和与差的正切公式(1)$$ \tan ( \alpha + \beta ) = \_ $$其中α,β$$ \alpha + \beta \neq $$$ k \pi + \frac { \pi } { 2 } ( k \in Z ) ; $$(2)$$ \tan ( \alpha - \beta ) = 其 $$其中α,β,$$ \alpha - \beta \neq $$$ k \pi + \frac...
( \alpha + \beta ) } $$的正切$$ k \pi + \frac { \pi } { 2 } ( k \in Z ) $$两角差$$ \tan ( \alpha - \beta ) \alpha , \beta , \alpha - \beta \neq $$=2$$ T _ { ( \alpha - \beta ) } $$的正切$$ k \pi + \frac { \pi } { 2 } ( k \in Z )...
解析 45{}^\circ 如图所示,\tan \angle BAE=\dfrac{3}{2},\tan \angle BAF=\dfrac{1}{5},∴\angle BAE=\alpha ,\angle BAF=\beta ,∴\alpha -\beta =\angle EAF,∵\triangle AEF是等腰直角三角形,∴\angle EAF=45{}^\circ ,∴\alpha -\beta =45{}^\circ . ...