Question Number 90739 by ajfour last updated on 25/Apr/20 $${solve}\:{for}\:{x}\:{and}\:{y} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right]+\mathrm{cot}\:^{\mathrm{2}} \left[\pi\left({x}+{y}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\sqrt{\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$ Commented by me2love2math last updated on…
Question Number 25198 by Joel578 last updated on 06/Dec/17 $$\mathrm{Given} \\ $$$$\mathrm{sin}\:\left({K}\:+\:{L}\right)\:\mathrm{cos}\:{M}\:=\:\mathrm{2}\:\mathrm{sin}\:{K}\:\mathrm{cos}\:\left({L}\:−\:{M}\right) \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{tan}\:{M}\:=\:\frac{\mathrm{sin}\:\left({L}\:−\:{K}\right)}{\mathrm{2}\:\mathrm{sin}\:{K}\:\mathrm{sin}\:{L}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 25191 by adityapratap2585@gmail.com last updated on 05/Dec/17 $$\mathrm{A}\:\mathrm{man}\:\mathrm{is}\:\mathrm{climbing}\:\mathrm{a}\:\mathrm{ladder}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{inclined}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{30}°\:. \\ $$$$\mathrm{If}\:\mathrm{he}\:\mathrm{ascends}\:\mathrm{at}\:\mathrm{a}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{2}\:\mathrm{m}/\mathrm{s}\:\mathrm{then}\: \\ $$$$\mathrm{he}\:\mathrm{approaches}\:\mathrm{the}\:\mathrm{wall}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}− \\ $$$$ \\ $$ Commented by prakash jain last…
Question Number 90700 by jagoll last updated on 25/Apr/20 $$\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{8}}\right)+\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)\:? \\ $$ Commented by john santu last updated on 25/Apr/20 $$\mathrm{sin}\:^{\mathrm{2}}…
Question Number 25148 by adityapratap2585@gmail.com last updated on 05/Dec/17 $$\mathrm{difference}\:\mathrm{between}\:\mathrm{degree}\:\mathrm{and}\:\mathrm{radian}. \\ $$ Commented by prakash jain last updated on 05/Dec/17 $$\mathrm{radian}\:\mathrm{is}\:\mathrm{measured}\:\mathrm{as}\:\mathrm{radio}\:\mathrm{of}\:\mathrm{arc} \\ $$$$\mathrm{length}\:\mathrm{to}\:\mathrm{radius}. \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{semicircle}\:\mathrm{arc}\:\mathrm{length}=\pi{r}…
Question Number 25125 by adityapratap2585@gmail.com last updated on 04/Dec/17 $$\boldsymbol{\mathrm{A}}\:\mathrm{vertical}\:\mathrm{stick}\:\mathrm{12}\:\boldsymbol{\mathrm{cm}}\:\mathrm{long}\:\mathrm{casts}\:\mathrm{a}\: \\ $$$$\mathrm{shadow}\:\mathrm{of}\:\mathrm{8}\:\boldsymbol{\mathrm{cm}}\:\mathrm{long}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}\:. \\ $$$$\mathrm{At}\:\mathrm{the}\:\mathrm{same}\:\mathrm{time},\:\mathrm{a}\:\mathrm{tower}\:\mathrm{casts}\:\mathrm{a}\: \\ $$$$\mathrm{shadow}\:\mathrm{of}\:\mathrm{40}\:\boldsymbol{\mathrm{m}}\:\mathrm{long}\:\mathrm{on}\:\mathrm{the}\:\mathrm{ground}.\: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{hight}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tower}\:? \\ $$ Commented by adityapratap2585@gmail.com last updated…
Question Number 90596 by I want to learn more last updated on 24/Apr/20 $$\mathrm{If}\:\:\:\mathrm{sin}\left(\mathrm{28}\right)\:\:=\:\:\mathrm{a}\:\:\:\:\mathrm{and}\:\:\:\mathrm{cos}\left(\mathrm{32}\right)\:\:=\:\:\mathrm{b} \\ $$$$\mathrm{Find}\:\:\left(\mathrm{i}\right)\:\:\mathrm{cos}\left(\mathrm{28}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{ii}\right)\:\mathrm{cos}\left(\mathrm{64}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{iii}\right)\:\mathrm{sin}\left(\mathrm{4}\right) \\ $$ Answered by mahdi…
If-a-sin-2-x-b-cos-2-x-c-b-sin-2-y-a-cos-2-y-d-and-a-tan-x-b-tan-y-then-a-2-b-2-in-terms-of-a-b-c-d-
Question Number 24973 by Rishabh#1 last updated on 30/Nov/17 $$\mathrm{If}\:{a}\:\mathrm{sin}^{\mathrm{2}} {x}+{b}\:\mathrm{cos}^{\mathrm{2}} {x}={c},\:{b}\:\mathrm{sin}^{\mathrm{2}} {y}+{a}\:\mathrm{cos}^{\mathrm{2}} {y}={d} \\ $$$$\mathrm{and}\:\:{a}\:\mathrm{tan}\:{x}=\:{b}\:\mathrm{tan}\:{y}\:\mathrm{then}\:\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=? \\ $$$$\left(\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{a},{b},{c},{d}\right) \\ $$ Commented by nnnavendu…
Question Number 24971 by Rishabh#1 last updated on 30/Nov/17 $$\mathrm{Show}\:\mathrm{that} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:\mathrm{3}{x}}+\frac{\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{cos}\:\mathrm{9}{x}}+\frac{\mathrm{cos}\:\mathrm{9}{x}}{\mathrm{cos}\:\mathrm{27}{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{tan}\:\mathrm{27}{x}−\mathrm{tan}\:{x}\right). \\ $$ Answered by math solver last updated on 09/Feb/18 $$\mathrm{assuming}\:\mathrm{3rd}\:\mathrm{term}\:\mathrm{to}\:\mathrm{be}\: \\ $$$$\:\:\Rightarrow\:\frac{\mathrm{sin9}{x}}{{cos}\mathrm{27}{x}}.…
Question Number 24972 by Rishabh#1 last updated on 30/Nov/17 $$\mathrm{If}\:\alpha,\beta\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{conected}\:\mathrm{by}\:\mathrm{the}\:\mathrm{relation} \\ $$$$\mathrm{2tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \alpha\:\mathrm{tan}^{\mathrm{2}} \beta\:+ \\ $$$$\:\:\:\mathrm{tan}^{\mathrm{2}} \beta\:\mathrm{tan}^{\mathrm{2}} \gamma+\mathrm{tan}^{\mathrm{2}} \gamma\:\mathrm{tan}^{\mathrm{2}} \alpha=\mathrm{1}\:\mathrm{then} \\ $$$$\mathrm{which}\:\mathrm{of}\:\mathrm{these}\:\mathrm{are}\:\mathrm{correct}\left(\mathrm{multi}\:\mathrm{correct}\right)…